כן, אוקיי. היי, everyone, thank you for coming. My name is Dov Hershkovitz. I'm a senior product manager at GitLab. This talk was mainly inspired by many conversations that I had with our customers and users, also some of the best practices that we've been implementing in GitLab. And hopefully by the end of this conversation, you'll better know how to secure your DevOps process using OpenStandard today without waiting for anyone to build any integration for you. Basically use the tools that you have today. So let's just start, but before we'll get going, a quick definition of what is a secret. A secret is an object that enables the process of accessing restricted data or a resource, meaning that the data itself or the environment that you want to access is not a secret, the data is protected, but the way to access the data, the key, is the actual secret. Now we all know that software is all about speed. Any software organization wants to develop and deliver features to our customers as quickly as possible. We want to react quickly, we want to close vulnerability holes and fix bugs as fast as we can. This is why DevOps becomes such a wide practice in our industry. It was widely adopted because it allows us to automate basically everything and speed up our process. But we also know that our pipelines and our software need to be secure. And sometimes it feels that security and the security priorities will reduce our cycle time and will not speed up our processes. And this is because security is more gate-driven, it's heavy on process, and let's be honest with ourselves, there is a lot of lack of knowledge around security because we don't know where the next vulnerability will come from, so we don't know which holes we need to close first, and there are always holes that we need to close. This is why DevSecOps as a concept had emerged, basically putting security at the heart of DevOps without compromising. Let's not compromise, let's optimize. Let's optimize speed, let's optimize security. Eventually what we'll gain is speedy and secure pipeline. So in order to achieve that, first of all, we need to understand that secrets need to live in a dedicated secret solution. This is why there are so many secret solutions out there. All the cloud providers have different secret management solutions and key management systems to store and handle secrets. And also there are the dedicated secret management solutions that you all know, like HashiCoVolt, Conju, Akiles, and more. There are a lot of tools in the market and a lot of solutions to choose from. But as we mentioned, we want to make sure that we are integrating this with our DevOps platform, meaning our CI-CD pipeline, because we want to pull those secrets and use that as a part of our processes to allow us to manage our infrastructure or deploy our application in a secure manner. And now we are facing with another dilemma. Which tool should we choose from? Should we simply choose our secret management solution based on the available integration? Doesn't make sense, we want to use the best of breed. What if my organization is using a tool that simply does not support my DevOps platform and cannot integrate me with my CI-CD pipeline? Or if I have a multi-cloud strategy and I want to integrate with all the different cloud providers? What should we do? What should you do? Well, if there is a native integration, if one of the vendors builds a native integration for you, then great, you can simply use it. But as a product manager in GitLab, as much as I would love to integrate with all the secret management solutions that are out there, I simply can't. So there is something that we can do until someone will build that integration for you. And to do that, we need to familiarize ourselves with two concepts, OpenID and the JWT. Now the OpenID is an open standard for decentralized authentication. The JWT is a token, one of the ways to implement OpenID Connect. It's a short lived ID token. And the good news is that all the cloud providers or the secret management solution and many of the DevOps platform support the OpenID with the JWT implementation. In fact, I still haven't seen the tool that does not support that. I think it's practically a standard today. The fact that it is a standard. So if we have that in our disposal, what can we do? The solution will be in the form of integrating our OpenID with our DevOps platform. So in order to do that, we need to do the following step, or you need to do the following steps. And it's practically the same, and it doesn't matter which tool you are choosing, but the steps are more or less the same. Number one, you need to configure trust relationship with OpenID collecting in a declarative manner. You need to tell your cloud providers or your secret management solution any service provider I want to authenticate using an OpenID Connect with a JWT token. And then any DevOps platform, or if it's Github, Github, CircleCI, all of them are capable of generating a JWT token and send it to your provider. Then it gets authenticated. And if it's get authenticated, you'll get an access token back. Once you get that access token back, you can go ahead and retrieve any secret that you like in order to implement it as a part of your CI-CD pipeline, as I mentioned, to manage your infrastructure or deploy your application. And that's it. This is how you set up an integration between any DevOps platform to any secret providers. Now, like anything in our industry, any solution in our industry, there are pros and cons for both. So let's work them quickly. The benefit of using the JWT in DevOps workflow is, first of all, it's way more secure because you're using a short-lived ID token. Even if the token gets leaked, it has an expiration time of five minutes, so there is probably not a lot of things that an attacker can do in a course of five minutes. You have no vendor look. You don't need to wait for anyone to build an integration for you. You can go and build that integration right now. And also, secret are stored and secure in a dedicated system. You gain all the features and all the benefit that the secret provider brings in. The downside of using the JWT in DevOps workflow is, first of all, you need to know how to configure OpenID Connect. And also, the setup or the configuration in your CI pipeline, also a little bit complex. I thought this is something that you need to do one time and then the entire organization can use. Still, someone needs to do it for you. Even in GitLab, we have, like, a handful of engineers that really know how to do it. And the rest of the engineers are simply using it. And also, there is an associate cost with storing secrets in cloud providers or any secret management solution. Lastly, if there is one slide that I want you to remember from this lighting talk, is this one basically recapping the entire presentation. The challenge, the way I see it, that secret are typically stored in a dedicated management solution. But we know that secrets need to integrate with our DevOps platform, meaning our CI-CD pipeline. The solution, in case you don't have a built-in integration, go ahead and configure OpenID Connect on your provider and set up your pipeline to use JWT to retrieve those secrets. And the result, the secrets are securely stored and integrated with your existing DevOps tools. Now, in this lighting talk, I don't have the time to go into the depth of how to build that integration, but I've created one slide with a lot of working examples that you can simply use. And it's very easy. I mean, all you need to do is go to your documentation and just look for OpenID Connect or OIDC and you'll immediately see it. With that, I want to thank you and hopefully you enjoyed this talk. Bye-bye.


A monopoly exists when a specific person or enterprise is the only supplier of a particular commodity. This contrasts with a monopsonate which relates to a single entity's control of the market to purchase a good or service, and with oligopoly which consists of a few sellers dominating a market. Two monopolies are thus characterized by a lack of economic competition to produce the good or service. A lack of viable substitute goods, and the possibility of a high monopoly price well above the seller's marginal cost that leads to a high monopoly profit.The verb monopolize or monopolize refers to the process by which the company gains the ability to raise prices or exclude competitors. In economics, a monopoly is a single seller. In law, a monopoly is a business entity that has significant market power, that is, the power to charge overly high prices. Although monopolies may be big businesses, size is not a characteristic of a monopoly. The small business may still have the power to raise prices in a small industry or market. A monopoly is distinguished from a monopsonate, in which there is only one buyer of a product or service, a monopoly may also have monopsonate control of a sector of a market. Likewise, a monopoly should be distinguished from a cartel of form of oligopoly in which several providers act together to coordinate services, prices or sale of goods. Monopsonates and oligopolys are all situations in which one or a few entities have market power and therefore interact with their customers. Monopoly or oligopoly or suppliers' monopsona in ways that distort the market. Monopolis can be established by a government, form naturally, or form by integration. In many jurisdictions, competition laws restrict monopolies. Holding the dominant position or a monopoly in a market is often not illegal in itself, however certain categories of behavior can be considered abusive and therefore incur legal sanctions when business is dominant. The government-granted monopoly or legal monopoly, by contrast, is sanctioned by the state, often to provide an incentive to invest in a risky venture or enrich a domestic interest group. Patents, copyrights, and trademarks are sometimes used as examples of government-granted monopolies. The government may also reserve the venture for itself, thus forming a government monopoly.citation needed. Monopolis may be naturally a current due to limited competition because the industry is resource-intensive and requires substantial costs to operate.


The Court of Appeal in Abuja has affirmed Julius Abouret as the Chairman of the Labour Party, LP. The Appellate Court also set aside the decision of the Federal Capital Territory, FCT High Court, which last year restrained Abouret and two others from parading themselves as national officers of the party. In a judgment delivered by Justice Hamar Barkha, the Appellate Court held that the High Court was wrong to have assumed jurisdiction on the matter. The Appellate Court, which also awarded a cost of one million Naira in favour of Abouret and against the respondent, Lamedi Apappa, held that Abouret's appeal has married and was thereby allowed. Following the 2023 presidential poll, with LP's Peter Obey coming 3rd, Abouret has been in a protracted legal conflict with Apappa, who laid claim to the party's chairmanship. The party has always been on the positive side that in any form of division, in the Labour Party, we have just been confirmed by the today's judgement. All we are saying is that our brothers out there should please come back. The party is open, the party is willing to read me them. Whatever the party had wanted was in New Nigeria and anyone that is desirous of the New Nigeria should please come and join Labour Party so that we can make this initial work again. We are not interested in a proposition within the House. The Appellate Court will be notified about fresh news updates.


Welcome to lecture 4 on measure and integration. I recall, we have been looking at classes of subsets of a set X with various properties. We will start with the collection called as semi-algebra of subsets of a set X. Then, we looked at what is called the algebra of subsets of a set X. Today, we will start with looking at some more classes of subsets of a set X. We will start with what is called a monotone class. Then, we will look at the monotone class generated by a collection of subsets of a set X. Then, we will go over to describe what is the monotone class generated by an algebra. That is an important relation, which we will be using again and again. Let us start with describing what is a monotone class. A monotone class is a collection of subsets of a set X. Let us denote that collection of subsets by M. M is a collection of subsets of X. We say it is a monotone class, if it has the following two properties 1. Whenever there is a sequence of sets A n belonging to M, the collection M says that the sequence is increasing. That means for every n, A n is a subset of A n plus 1. Then, we demand that the union of these sets A n's also belongs to M. The first property is that the collection M of subsets of set X is closed under unions of increasing sequences. The second property that we expect from this collection is that whenever a sequence A n is in M and A n is decreasing, that means A n is a subset of A n plus 1 for every n, then the intersection of this sequence of sets A n should also be an element of M. Let us recall once again what is a monotone class. A monotone class is a collection of subsets of a set X with the two properties 1. For every sequence of sets A n in M such that A n's is increasing, their union also belongs to M. Secondly, whenever A n is a sequence of sets in M such that A n is decreasing, then the intersection of these sets is also in M. That is why the name monotone comes. That means this class M of subsets of M is closed under monotone sequences. Whenever a sequence A n is increasing in M, the union belongs to M. Whenever a sequence A n is decreasing in M, then their intersection also belongs to M. Such a collection of subsets of a set X is called a monotone class. Let us look at some examples of such collections. Firstly, let us observe every sigma algebra is also a monotone class. Why is that true? Because a sigma algebra is a collection of subsets of X which is closed under any countable unions. Because it is closed under countable unions, it will also be closed under increasing unions. First property will be true. Secondly, if a sequence A n is decreasing in a sigma algebra, then look at the complements of that sequence. So, that sequence of complements of these sets will be an increasing sequence of sets. Because it is a sigma algebra, it is also closed under complements. So, A n complements will belong to it. So, union of A n complements belongs to it. That means the intersections of A n's complements belong to it and that means the intersections of A n's belong to it. Let us look at this property. How do we write it? Suppose M is a sigma algebra. Claim M is a monotone class. To prove this, how do we go ahead? So, one, let A n belong to M and A n's increase. So, that is A n is a subset of A n plus 1 for every n bigger than or equal to 1. Then, union of A n's n equal to 1 to infinity belong to M because M is a sigma algebra. Because it is a sigma algebra, it is closed under all unions and n's such types also. Secondly, let us take let A n belong to M and A n's decrease. So, that is A n includes A n plus 1 for every n bigger than or equal to 1. So, in that case, that implies because A n's belong to M and M is so A n complements belong to M for every n bigger than or equal to 1 because M is a sigma algebra. Because it is a sigma algebra, it is closed under a complements. So, once that is true, that implies that A n complements union n equal to 1 to infinity belongs to M. Because A n decreasing implies the sequence of A n complements will be increasing. Just now, we saw that whenever a sequence is increasing, their unions belong to M. But that implies, what is this set? That is intersection of n equal to 1 to infinity A n complements. That is by De Morgan's law. So, this belongs to M and now M is a sigma algebra. So, that implies that intersection n equal to 1 to infinity A n belongs to M. So, we have shown whenever sequence A n is in M and A n's are decreasing, that implies the intersection also belongs to M. So, hence M is a monotone class. So, the first proposition or first observation is that every sigma algebra is also a monotone class. Let us go over to some more properties, more examples. Let X be any uncountable set and let us look at the collection of all subsets A of X, say that A is a countable set. The claim is that this collection M is a monotone class, but it is not a sigma algebra. So, let us look at M. X is uncountable and we are looking at the collection M of all those subsets of X, such that A is countable. So, claim M is a monotone class. So, let us see how do we prove it. So, first property. So, let us take a sequence A n belonging to M, A n's increasing A n subset of A n plus 1 for every n. What we have to prove? We have to check that union of A n's n equal to 1 to infinity also belongs to M. That is what we have to check. But let us note, to check that the union belongs to M, we have to show it is a countable set. Now, each A n is given to be an element of M. That means each A n is countable, A n countable for every n implies union A n is also countable. So, here we are using the effect that the countable union of countable sets is again a countable set. Hence, this implies that union A n's n equal to 1 to infinity belongs to M. The first property we have checked is that if A n's belong to M and A n's is an increasing sequence, then the union A n belongs to M. Let us check the second property namely, let us check the second property that A n belonging to M. A n's decreasing for every n should imply that the intersection A n's belongs to M. What we have to check? We have to check that intersection A n, n equal to 1 to infinity belongs to M. That means it is countable. That is what we have to check. But let us observe that intersection A n's, this is a subset of A n for every n because it is intersection and each A n is countable, A n countable implies and this is a subset of it. So, intersection A n is countable and hence implies intersection A n belongs to M. We have shown that if you take the collection of, if X is a countable or uncountable set does not matter actually, so far what we have we have not used the fact it is a uncountable set. If M is a set which is, if M is the collection of all countable subsets of a set X, then that forms a monotone class. Why it is not a, so claim, so finally we want to prove that M is not a sigma algebra. Let us observe a few things here. So, note here we will be using X uncountable implies first of all X does not belong to, X does not belong to M. So, the very first property of a collection being a sigma algebra, namely the whole space belong to it is violated because X is not countable it is uncountable. Another way of looking at this is the following. So, this is one observation. Secondly, X uncountable implies there exists a subset A in X such that neither a nor a complement is countable and that is obvious because if let us see is that X is a uncountable set then the claim that there exists, this is not actually required and this may not be true. Let us take a subset A of X say that either A or A complement is not countable and that is possible because if for every set A and A complement are countable then X will be a countable set. So, let us choose a set A say that either A or A complement is not countable then suppose A is countable A is countable. So, that will imply that A belongs to M, but A complement is not countable and that implies A complement does not belong to M. So, when X is uncountable you can have in fact for every set which is countable A will belong to M, but A complement will not belong to M. So, this collection M is also not going to be closed under complements. So, X does not belong to it and that it is also not going to be closed under complements and that is because X is uncountable. So, when X is uncountable collection M of all countable subsets of it is a monotone class, but it is not a sigma algebra. So, every sigma algebra is a monotone class, but the converse need not be true. So, this is what we have shown just now that if M is a monotone class, every monotone class is a sigma algebra, but there exist examples of sigma algebra, examples of monotone classes which are not sigma algebra. Let us look at the next scenario. Let us start with a collection C of subsets of a set X. C is any collection it may or may not be a monotone class. So, in we would like to find a monotone class of subsets of X which includes C and is smallest. So, first of all let us observe that all subsets of the set X is a monotone class of subsets of X and C is because it is a subset is a sub collection. So, C is a sub collection. So, given any collection of subsets of a set X C it is always included in a collection namely the power set of X which is a monotone class. So, given a collection there always exists a monotone class of subsets of X including it, but this is too large we want to have the smallest monotone class including C whether such a thing exists or not. So, the proof is something similar to what we have shown is algebra generated by a class, the sigma algebra generated by a class. So, let us look at M of C. So, this is a notation for the intersection of all monotone classes M of subsets of X which includes C. So, look at the collection of all monotone classes of subsets of X which includes C and take their intersection and call this as M of C. So, what we want to prove is that M of C is a monotone class it includes C and it is the smallest. So, let us look at a proof of that. So, C is any collection of subsets of X, M of C is the intersection of all monotone classes M, M including C. Claim 1, C is inside M of C that is obvious because C is inside every collection M and M of C is the intersection. So, this property is obvious. Second, claim that M of C is a monotone class and the proof is goes on the same lines as that of algebra and semi-algebra, algebra and sigma algebra. To prove this, let us look at A ns belong to M of C, A ns decreasing. But this implies that each A n also belongs to M, also belongs to M for every collection M which includes C for every n and that implies that union of is decreasing. So, let us we want to show that intersection A ns n equal to 1 to infinity belong to M for every m and that implies that M is every m and hence that implies that intersection A ns belong to M of C. So, essentially saying that if A n is a sequence in M of C which is decreasing then it is also a sequence which is decreasing in each m and hence the intersection belongs to each m and hence belongs to the intersection M of C. And the same proof, similar proof works for the union. So, let us look at the second part A ns belong to M of C and A ns increasing but that means implies A ns belong to M for every m for every m and A n is increasing that means A ns union because each m is a monotone class. So, this belongs to M, the union belongs to M for every m and that implies that the union A ns n equal to 1 to infinity belongs to M of C and that implies. So, hence M of C is a monotone class. So, it is a monotone class that includes the collection C and third M of C is smallest such that C is M of C, smallest monotone class and that is obvious because it is the intersection of all monotone classes. So, obviously it is going to be the smallest. So, what we have shown is that given a collection C of subsets of a set X there exists a monotone class M of C which includes C and which is the smallest. So, this monotone class is called. So, this is what we have proved. So, thus M of C is the smallest monotone class of subsets of X such that C is the entire M of C and this collection is called the monotone class generated by C. So, every collection C has got the smallest monotone class in which includes it. So, that is called the monotone class generated by it. So, given a collection C of subsets of a set X we are able to generate an algebra out of it. We are able to generate a monotone class out of it. We are able to generate a sigma algebra out of it and the next question that we want to analyze is what is the relation between these collections. So, we want to prove a theorem which relates these concepts. So, first of all let us start with any collection of subsets of a set X. Then the first observation is if C is an algebra which is also a monotone class then C is a sigma algebra. So, let us first prove this fact that if C is C an algebra plus monotone class implies C is a sigma algebra. So, what we have to prove? To prove C is a sigma algebra, we have to prove first empty set the whole space belongs to C. That is true because C is an algebra. So, this property is true. Second, we should show that if a set A belongs to C implies A complement belongs to C and that is again obvious because it is the collection C is a algebra. So, these two properties are true. The third property is the only property to be checked that if A n belongs to C then that should imply union of A n is n equal to 1 to infinity also belongs to C. So, C is closed under countable unions that is what we have to prove and what we are given is C is a monotone class. So, a monotone class is a collection which is closed under only increasing and decreasing. So, let us look at union of A n's. Can we represent this as a union of increasing sets? Well, the one possibility is let us take union of A i's i equal to 1 to n then this collection as n increases will be an increasing sequence and the union n equal to 1 to infinity will be an increasing union. It will be a union of increasing sequence of sets which are namely union A i's and now observe that each one of these sets union A i i equal to 1 to n that is a finite union of elements in the algebra. A n's belong to C and that is an algebra. So, that means each one of them belong to C. So, we have written the union A n's as union of sets in C and this is an increasing sequence and C is a monotone class. So, that implies that this right hand side this set belongs to C. So, note we have used we have represented any union as a increasing as a union of increasing sequence of sets and each set here is a finite union of elements of the algebra C. So, hence this belongs to it. So, this becomes an increasing union of sets and hence because it is a monotone class this belongs to C. So, every A n belonging to C implies the union 1 to infinity also belongs to C. In fact, it is closed under countable unions. So, it becomes a sigma algebra. So, this proves the first property namely if C is an algebra which is also a monotone class then it is a sigma algebra. Let us look at the next property that if C is any collection then C is contained in M of C. That is obvious because M of C is the smallest monotone class including C. So, this is obvious and now C is also subset of S of C because S of C is the sigma algebra generated by it. So, C is inside S of C and just now we proved S of C is also a monotone class. So, S of C is a monotone class including C and hence the smallest one must come inside. So, that we prove this C is inside M of C is inside S of C. That means given any collection of subsets of x it is always included in the monotone class generated by it and the monotone class generated by it is also inside the sigma algebra generated by it. So, let me repeat these arguments once again. So, first of all C is contained in M of C that is because M of C is the smallest monotone class including C. Also, C is included in S of C because S of C is the smallest sigma algebra of subsets of x which includes C and this just now we proved is also a monotone class. So, this is a monotone class including C. So, that implies the smallest one must come inside it and the smallest one is M of C that comes inside S of C. So, what we have shown is for every collection C of subsets of x, the monotone class generated by it is a subset of the sigma algebra generated by it. We want to analyze the question when can we say S of C is also a subset of M of C. When is this true? That is same as saying that when is S of C the sigma algebra generated by a collection can I say it is equal to M of C. The answer is given by the next theorem which says that if C is an algebra then this is true. So, this is an important theorem called the sigma algebra monotone class theorem. It says if A is an algebra of subsets of a set X then the sigma algebra generated by it is same as the monotone class generated by it. Now, we have observed that M of, so the first part we have already observed that M of A is a subset of S of A. For that one need not have even A as algebra for any collection M of C is contained in S of C. So, in particular if A is an algebra then M of A is a subset of S of A. So, we have to prove the second the other way around inclusion namely M of A includes S of A when A is an algebra. To prove the other way around inclusion let us observe it is enough to prove that M of A is an algebra because why it is enough to prove this? Let us observe enough to show M of A is an algebra will imply because M of A is an algebra also a monotone class and just now we proved every algebra which is a monotone class will imply M of A is a sigma algebra because M of A is a monotone class and if we are able to show it is an algebra then it will be also a sigma algebra. Then we have A is inside M of A and now if M of A is a sigma algebra that even imply the smallest one must come inside it. So, S of A will be inside M of A. So, that will prove S of A is a subset of M of A and will be through. So, we have to only prove to show M of A is an algebra when A is an algebra. So, this is what we have to prove. So, let us start looking at a proof of this. So, first of all to prove that M of A is an algebra we should show that it is closed under complements. So, let us try to prove it is closed under complements. So, that means what to show it is closed under complements I have to show that for every subset in M of A its complement is also in M of A. So, this is a technique which we are going to use very often. So, let us collect together all the sets B which have the property that whenever B is the collection of all those subsets say that E complement belongs to M of A. So, to prove M of A is closed under complements what we have to show is that M of A is a subset of B. So, M of A is a subset of B we have to show that. So, let us try to prove that. So, claim let me repeat that we want to show that M of A is closed under the operation of complements that is A belonging to M of A should imply A complement belong to M of A. So, to show that consider all those subsets. So, let us consider the collection B of all those subsets E contained in X such that E complement belongs to M of A. So, to prove this this will be true. So, the required claim will be true. So, let us so this will be true if we can show M of A is contained in B. So, that is what we want to prove. We want to show because then for every set A belonging to M of A it will belong to B that means its complement will belong to M of A. So, this is what we have to show. Now, let us observe we are trying to show that M of A is inside a collection B and what is M of A? M of A is the smallest monotone class including A. So, suppose we are able to show that A is inside B and B is a monotone class then this claim will be true. So, for this enough to show. So, to prove this claim it is enough to show one that A is inside B and second we should show that M of and secondly B is a monotone class because once B is a monotone class including A the smallest one will come inside. So, let us try to prove these two facts that A is so proof of one that A is a subset of B. So, let set A belongs to A that implies A is algebra. So, that implies A complement belongs to A because A is algebra and note implies A complement belongs to A which is inside M of A because A is always inside M of A. So, what we have shown that if A belongs to A then its complement belongs to M of A. Hence that is same as saying that A belongs to the collection B. So, we have proved that B includes A. So, first property is true. Let us look at the second property. So, what is the second property we want to prove? The second property we want to prove is that B is a monotone class. So, let us take a collection A n, a sequence belonging to B such that it is decreasing or increasing. So, let us say A n is increasing. But A n belonging to B means what? That means what A n complements belong to M of A. That is the definition of the class B. So, saying that we have got a set A n in this that means A n complements belong to A. Now, M of A is a monotone class. So, that implies that A n complements intersection will belong to M of A. Provided we can say A n complements are decreasing and that is true because A n's are increasing because A n complements are decreasing and M of A is a monotone class. So, that means this intersection belongs to it. So, that means union of A n's n equal to 1 to infinity, complement of this belongs to M of A. So, whenever a sequence A n belongs to B and A n's are increasing, we have got the complement union. The complement of the union belongs to it. So, that means union of A n's n equal to 1 to infinity belongs to B. So, the collection B is closed under increasing unions and let us finally prove that it is also closed under decreasing sequences. .. Let us take a sequence of sets which is decreasing. So, let A n belong to B and A n's decrease. We want to show that the intersection of A n's belong to it, but A n's belong to B implies that just now we observed A n by definition A n complements belong to M of A. By definition A n belongs to B means A n complements belong to A for every n and that implies now A n complements is a sequence because A n's are decreasing that is same as A n complements are increasing. So, union of A n complements belong to M of A because M of A is a monotone class and that implies that if the intersection of A n's n equal to 1 to infinity complement belongs to M of A. So, whenever A n's belong to it and take the intersection of A n's there complement belong to it that means intersection of A n, n equal to 1 to infinity belongs to B. So, what we have shown is the collection B is closed under increasing unions is closed under decreasing intersections that means B is a monotone class. So, A is inside B, B is a monotone class and that will prove that M of A is a subset of B because this is a monotone class including A. So, it must include the smallest one. So, that proves the first step of our claim namely that the collection M of A is closed under complements. We wanted to show it is a algebra. So, what is the next step? Next step should be to show that M of A is closed under unions. So, that means whenever two sets E and F belong to M of A the union must belong to M of A. So, let us fix one of them. Let us fix the set F in M of A and let us look at the collection L of F such that it is a collection of all those sets say that A union F belongs to M of A. So, what we have to prove? So, in this we have to prove that M of A is closed under unions. We have to prove that M of A is a subset of L of F. So, once again the required property that M of A is closed under unions we are translating into a property of a collection of subsets. So, let us try to show that M of A is contained in L of F. So, that is first we should try to show. So, to show that the collection M of A is closed under unions. So, this is what we want to show. So, let us fix a set F belonging to M of A and consider the collection which is let us call it L of F. What is this collection? It is a collection of all those subsets in X such that E union F belongs to M of A. So, saying that M of A is closed under unions so to show. So, we should show that M of A is a subset of L of F. So, that is what we should show. So, once again we want to show M of A is a subset of L of F. M of A is a monotone class generated by A and we want to show it comes under L of F in some other collection L of F. So, that means we should try to show that A is inside this collection and this collection L of F is a monotone class. So, we should try to show that A is inside L of F for every F belonging to M of A and second L of F is a monotone class. Let us just observe second one which is quite obvious. So, let us observe L of F is a monotone class. So, for that what we have to show let us take a sequence En belonging to L of F, En's increasing. But that will mean if En's are in L of F that will imply that En union F belongs to M of A. That is by definition of L of F and M of A is a monotone class. En's are increasing. So, En union F is also increasing. So, implies that union of En union F belongs to M of A because En's are increasing. En union F is increasing and belongs and that means it is same as saying that union of En's union F belongs to M of A and what does that mean? That means union of En's belong to the class L of F. So, whenever En's belong to L of F, En's increasing this implies that union En's belong to L of F. So, this is what we have just now proved. A similar proof will work for decreasing also. So, saying that L of F is a monotone class is a straightforward argument because M of A is a monotone class. Let us try to check that A is inside L of F for every F belonging to L of M of A. So, we want to check the first property namely. So, this is a property we want to check that this collection algebra is inside L of F for every F belonging to M of A. Let us note for the time being, we want to check this property for every F in M of A. Note if F belongs to A, then for every E belonging to A, E union F belongs to A because A is a algebra. If E and F are two sets in A, A is algebra that will mean that the union belongs to algebra and that is included in M of A. So, what does this imply? This means for every F in A, E union F belongs to M of A. That means that the set E belongs to the collection L of F. Once again, we are starting with a very simple observation. If a set F belongs to A and E belongs to A, then the union belongs to A and A is always inside M of A. So, that means E union F belongs to M of A for every E belonging to A. That means for every E belonging to A. So, that means we have shown that A is inside L of F. Now, A is inside L of F for every F in A implies L of F is a monotone class. This now proved implies M of A is inside L of F for every F belonging to A. So, what we have shown is M of A is a subset of L of F for every F belonging to A, but we want to check that M of A is inside F for every F belonging to M of A. We have got only for F belonging to A. Now, here is a very simple observation which helps us note a set E belongs to L of F if and only if F belongs to L of E. So, this is an observation which is going to be very important and very useful for us. A set E belongs to L of F means what? E union F belongs to M of A, but if E union F belongs to M of A, that is same as F union E belongs to that means F belongs to L of E. So, saying that E belongs to L of F is same as F belongs to L of A. Now, let us translate this property. Here, it says M of A is inside L of F for every F belonging to A. So, that means for every E belonging to M of A implies that E belongs to L of F and that is if and only if F belongs to L of E. Here F was in the algebra, F in the algebra. So, what we have got? For every F in the algebra L, it belongs to L of E whenever E belongs to L of A. So, that means what? That means A is inside L of F for every F belonging to M of A. See, how nicely we have turned the tables. Earlier we had M of A is inside L of F for every F in A. So, that means every element here E is element in L of F, but here E belongs to L of F means F belongs to L of E. Now, F is in A that means A is in L of F for every F belonging to A, but that means once that is true, now A is inside L of F for every F in M of A and that implies that M of A is inside L of F for every F belonging to M of A. Because L of F is a monodon class, it includes the algebra A. So, it must include the smallest one. So, M of A is inside L of F for every F belonging to M of A. So, we have proved the required thing, namely M of A is inside L of F for every F belonging to M of A and hence that means M of A is also closed under. So, here is what we wanted to prove. So, M of A is closed under unions that we translated into the property that M of A is inside L of F for every F in M of A and that we just now proved finally and you see again and again whenever we want to show something is true, we convert it into a property of a collection of objects and show generators come inside and everything comes inside. So, that proves that M of A is an algebra, M of A is an algebra, it is already a monodon class. So, it must be a sigma algebra and S of A is a sigma algebra. So, that will prove that the required theorem. So, this step proves the required theorem namely. So, for this, let us just go through this proof again to show that M of A is closed under unions. We fix F in M of A and look at this collection and we want to show M of A is inside L of F. For this, we have to show that L of F is a monodon class and A is inside L of F whenever F belongs to A. So, that says M of A will come inside for F belonging to A and now reverse the roles of these two that M of A for F in A that means E belongs to L of F. So, F belongs to L of E and reverse the roles and that comes. So, that gives you the property that M of A is inside L of F. So, it is closed under unions and hence it is a sigma algebra. So, it must include the smallest one and that proves the fact that the sigma algebra generated by algebra is same as the monodon class generated by the algebra. So, let us just recall what we have done till now. We started with this at X, looked at a collection C of subsets C contained in Px. The first thing we have looked at what is called a semi-algebra. Then, we looked at this collection C to be an algebra. Then, we looked at this collection to be a sigma algebra and then it to be a monodon class. So, a semi-algebra, every algebra is a semi-algebra, every sigma algebra is also an algebra. Every sigma algebra is also a monodon class. So, this is something here. This way around implication may not be true, this way around implication may not be true and this way around implication may not be true. Finally, we proved monodon class generated by an algebra is the sigma algebra generated by algebra, if A is an algebra. So, that finishes our study of collections of subsets of X with special properties. I just want to leave you with some exercises which you should try, which are important. The first property or first exercise we would like to, that whenever a collection is an algebra, it is equivalent to saying that empty set in the whole space, the collection is closed under complements, it is closed under unions and that is equivalent to saying whether it is closed under complements because of this de Morgan laws. Another property that whenever something is an algebra, a collection algebra, it is also closed under symmetric references and any finite union in an algebra can be represented as a finite disjoint union whenever you are inside algebra. That property we have seen, but you should try to prove this exercise yourself. Another property about semi-algebras is you can or sigma-algebras is that you can restrict. So, take a collection C of subsets of a set X and restrict it to a set E. That means, take intersection of all sets in C with E and this is C restricted to E and the property we want to prove is that if C is a semi-algebra, then C intersection E is a semi-algebra of subsets of E and similarly, you prove the property that the algebra generated by the restricted sets is same as the algebra generated by generate the algebra and restrict. So, f of C intersection E is equivalent to f of C intersection E. So, you restrict and generate, it is same as generate and restrict and the same property is true for algebras is true for sigma-algebras. So, that is the property about sigma-algebras and these exercises you should try to prove yourself. The steps are outlined here for you to prove. We have already proved these things in our lectures, but I will strongly advise that you prove these things yourself. Here is another example of generating new sigma-algebras, semi-algebras. F is a function from x to y. If you take sets in y and take inverse images that give you a collection of subsets of x. So, try to show that whenever if C is a collection of subsets of y, which is a semi-algebra or a sigma-algebra, then the pullback sets also form a semi-algebra or sigma-algebra. Here is another example of taking two collections of subsets f and g of two sets x and y. Look at the addition products of these collections show that in general it is not a algebra.  So, these properties unions may not be true. So, show that for union this property need not be true. So, that sigma-algebra technique, that same thing is inside, then the generated sigma-algebra comes inside that we use. So, we use that to prove that if you take the collection of all intervals in i and left open or right closed intervals, then the sigma-algebra generated by all intervals is same as the sigma-algebra generated by all left open right closed intervals and same as the boron sigma-algebra. So, I would strongly advise you to try these properties to get used to this concept of algebra, semi-algebra, sigma-algebra and monotone class. So, let us stop here today. Thank you very much.


Ladies and gentlemen, the story you are about to hear is true. The names have been changed to protect the innocent.  You're assigned a juvenile detail.  The articles taken and the method of operation indicate a juvenile thief. Your job, find him. It was Tuesday, July 6th. It was hot in Los Angeles. We were working the day watch out at juvenile details. My partner is Frank Smith. The boss is Captain Powers. My name's Friday. We're on our way out from the office and it was 10.36 a.m. when we got to the old Central Jail building. Crime lab. Lady. Probably will, too. Any water around? A bottle over there in the corner. Help yourself. Thanks. How about you, Joe? No, no, thanks. You got anything for us, Ray? Not the same as usual. Take a look. Same entrance. Cut the screen on one of the back windows, broke the glass and climbed in. Any friends? None of them could find. Here's a picture of the edge of the screen. Here. You can see the cut edges. You got anything from it? Not much. The figure was done with a knife.  What about the physical evidence in the house, Ray? Not much. We've got a couple of partials. Lifted them from the glass top on the dresser. Here. You can see here, not much is going. Match up with the prints from the other jobs? Couple of points, but not enough to go the case on. The only thing you got that definitely ties this one in with the other is the footprint. Another one? Mm-hmm. Found it in the soft earth underneath the window. Hey, man, Frank, what do you think? Oh, yeah. Here you go. Here. You can see it clearly, the heel mark. Same as in the other. Did you get a cast of it? Mm-hmm. You want to see it? No, no reason if you've looked it over. You got the same thing on the picture. Anything else? You had it. You got a cigarette, ran out and had a chance to get some more. Here. Thanks. Thanks, Joe. Yeah. You've been able to taste down the rubber heel for us? No, no luck. We contacted half a dozen firms. They all tell us the same things, made special. Probably some small factory specialized in custom-made shoes. Might be some sort of correcting shoe. You know, I think... Oh, yeah. You've had it for a couple of weeks now. What do you sit on it? I don't know where we found it. We talked about every repairman in town. None of them have seen this kind of a heel before. None of them can tell us where it was made. Mm-hmm. Check with the doctors in the area. None of them know the brand. I figured it might be one of them who sent for it. That's none of them. No. It's a new one to them, too. Yeah. Which is more, I could give you some. Well, that's all right, man. It seems no matter which way we turn, there's another wall waiting for us. How about Berger? Anything coming from them? Well, they're checking it all down. They got the same thing we came up with. Anything turn up on the stolen merchandise? No. Berger, he's made all the checks. Pawn shops have been alerted. Stuff's all cheap costume jewelry. No reason to try and hawk it. They wouldn't get anything for it. Mm-hmm. Brings it back to a motive. Why is he pulling the jabs in the first place? All right. We'll file that with where the heels are made. The only thing we can figure is that some kid out for kicks gets his laugh from breaking into houses. There's no other reason. Stuff he's stolen, you can buy it by the carload in any dime store. A couple of times, he's left money alone at some plain site to steal some cheap pin. Looks like you've got yourself a wild one. Yeah. Well, what can you tell us about the foot gun, sonny? Not much. Found under the window. Kid wasn't moving when they were made. Can't figure a stride. Mm-hmm. What about weight? Only a guess, Joe. Yeah. I'd say it came in about 140. Could be 15 pounds one way or the other. What about the size? Seven. Looks like seven C. Mm-hmm. You sure about that? Well, pretty sure. Not much, is it? Best we can do. Okay. Last place, a woman with real sore. Don't blame her. Not about the costume jewelry. She didn't much care about that. Surprise gladiola she was raising. Get tramped all over him. Real sore that he stepped on. Don't blame her. Says when she catches inches, she's gonna give him a good talking to. Tramping up all her plants. Well, we'll see you, Ray. Yeah. Sure haven't got much to go on, have ya? You called it. Frank and I went back to the office and we pulled all of the reports that we had on the thefts. We spent the rest of the morning going over them. The method of operation was apparent. Each Friday night, for the past two months, between the hours of 8.30 and 11 p.m., a single thief had entered a home in a wealthy section of town and stolen as much worthless costume jewelry as he could carry out. On each occasion, the entry was made through a rear window. In those cases where the window was locked, the thief broke the glass and undid the latch. On each occasion, he'd left at least one footprint in the dirt around the house. The heel impression was of a type that we hadn't been able to identify. We'd gotten pictures of it out to all major shoe manufacturers, but they were unable to tell us where the shoes had been made. We'd canvassed the area where the thefts were taking place without results. Rolling stake-outs were set up on Friday nights and they'd netted us nothing. The thief approached the home, broke into him, stole what he wanted, and then he left without being apprehended or seen. Throughout the investigation, Frank and I had been working closely with burglary division, but their men hadn't been able to come up with any more than we had. At 1.15 p.m., Frank and I left the office. We got something to eat and then we started to check the playgrounds again. At each one, we'd contact the member of the park department and asked them if they'd seen anyone wearing a pair of shoes similar to the pictures that we had. Every place we got the same answer, none of them had. We stopped groups of boys on the street and asked to see their shoes. We checked mall shops, drugstores, pool halls, and bowling alleys without any result. The same operation had been going on for the past two weeks without bettering our chances of apprehending the suspect. 5.30 p.m. Frank and I left one of the larger bowling alleys in town and headed back to the office. Boy, I'll sure be glad to get home tonight. Yeah. Hot. Sure was. Yeah, I must have gone over 90. Yeah. Must have. Paper said 90, but it was hotter, I can always tell. How can you always tell? My feet, Joe. Always tell them feet. Cold day, I can walk for a couple of miles and not have any trouble. Come the hot weather and if I step off a curb, I got misery. Yeah. I think I'll get me another pair of shoes, kind with the holes in them. What? Kind with the holes in them. You know what I mean? You see them all the time in the men's magazine kind of cloth on the top. You know, real thin leather on the sole and cloth over the top, a lot of little holes in it. They're cool. Cool. See them all the time in the magazine, don't you? Oh, yeah. Yeah, I see them all the time.  Always got on the same suit and he's got a straw hat with one of those bands made out of feathers, you know, like a kind of snake kind of... Mm-hmm. Guy's always in a tan suit and a hat. Maybe you gotta have one of those before you can buy the shoes, you figure? One of those suits? Yeah. I don't know. You're not talking much today. I haven't got nothing to get in. You're doing it all. You know? Oh, hard. That's pretty funny. It's hot. Must have been overnighting. I was just about to say it. Slun on it. Slun on it. Huh? Take it easy. You see a kid up there? No. Up there. Just coming out of the drugstore in the corner, waiting for the light there. Oh, yeah. And that's player, huh? At least wearing a pair of canvas shoes looks like from here. Want to check them? You might as well. That's good here. All right. You want to wait? No, I'll come with you. Hey, fella. Hold it up. What? Wait a minute. You mean me? Yeah. Yeah, what do you want? Police officers. No, we didn't say you did. We just want to ask a couple questions. Oh, I sat there for a minute. You were after me. My partner, Frank Smith, my name is Friday. Hi. Oh, I'm Alec McCarran. What do you want to ask? You been playing tennis? No, just practicing. I go down to the handball court at school and practice. I couldn't get a game this afternoon. Why? Let's take a look at your shoes, will you? What?  We want to see them. What is it? It's a joke? No. Well, it sure sounds like a rib. You guys sure you're really cop? Well, here's the ID again. You want to look? Okay, no reason not to, there. The souls, we want to see. You sure this isn't some kind of joke? Looks like no joke. Come on, Alec. Let's see the souls. Sure, there. Well, are you satisfied now? Yeah. Okay for me to go now? Afraid not. Well, I say, what is this anyway? Oh, it's stopping me in the shoe thing. You looking for something special? Not anymore. We just found it. Frank and I took the boy back to Georgia Street and checked his name through R&I. We found that there was no previous arrest record for him. It was at 7.15 p.m. We took him into one of the interrogation rooms. Your full name's Alec McCarran, is that right? Yeah. How old are you? 16. Where do you live? 4782 Fargo Street. Where's that son? Up above Silver Lake. You live with your family? Yeah. Mother and father, both living? Yeah. When are you going to tell me what this is all about? You got a job, son? Not now. I got one coming up. Doing what? Box boy in a market. Supposed to be to work next week. Kid they got now is going to the Army. They promised me his job. You go to school? Yeah, I'm a senior in high school. Look, it's getting late. My dad's going to start to worry. Is it okay if I call him? We'll call him for you in a minute. Thanks. What do I tell him when he wants to know what's happening? We'll talk to him. Okay. What do you usually do nights? I don't know what you mean. Well, evening hours. What do you do? I stay home mostly. I got to get my grades up for college. I usually study. During summer vacation? My dad helps me. I got to make up a couple of credits. I thought you had to go to summer school for that. I guess I'll get out of just about, yeah. But not everyone, though. No, I go out sometimes. When? Mostly the weekends, Friday. What do you do when you go out? Where do you go? Show sometimes. Anywhere else? Bowling. Sometimes just down at a library to get a book. Mostly to the show. Do you go with anybody? No. Where'd you get those tennis shoes? My dad got them for me. Where? Place back east. I had them a long time. You know the name of the place? Not right off. Some place in Maine, I think. You know anybody else who's got a pair like them? I've never seen any others. I told you my father got them for me. He used to wear them when he was a kid, sent back to Maine for them. Mm-hmm.  Yeah, we will. There've been a string of burglaries in town, housebreaking. A person that's been pulling them wears a pair of shoes just like yours. Like mine? That's right. You think I'm the one? Well, it looks like it, doesn't it? I'd better call my father. All right, we'll call him for you. He's the only pair of shoes in the world, you know. Only ones we've been able to find, son. There must be a lot more. We're willing to see him. I bet he lost answers to the police department. Who am I speaking to, please? Yes, sir. Your son's here. A suspicion of burglaries. That's right. Well, that may be so, Mr. McCarran, but that's the charge. Yes, sir? No, no. We'll bring him to his home. Sure, you can. Just a minute. I want to talk to your son. Hello, Dad? Yeah. Uh-huh. At the police station. No, it's no joke. What? Well, I'll tell you when I see you. Yeah, bye. All right, I'll get you some later. You taking me home now? That's right. Because Dad said you're straight, huh? Told you it was all a mistake? No, I'm afraid he didn't. What do you mean?  Well, he tried to, yeah. That's right. Well, then you're going to have to let me go. No, I'm afraid not. But you have to if it's a mistake. Yeah, if it's a mistake. Frank and I left the office with the boy and we drove out to his home. It was a large two-story colonial house overlooking Silver Lake. We were admitted to the place by a man we were shown into the living room. The boy's father, Henry McCarron, was waiting for us. You're the officer? Yes, that's right. My partner, Frank Smith, my name's Friday. Oh, yeah, yeah. You want to tell me what this is all about now? Well, it looks like your son's mixed up in some burglaries, Mr. McCarron. You're aware that's a pretty serious charge. Yes, that's right. I suppose you have some kind of proof to make that statement? We think so. Am I telling you what it is? Well, while we were talking, would it be all right to look at the boy's room? Do you think that's quite fair? Necessary, Mr. McCarron. What if I say you can't? Well, then we'll have to come back with a warrant. All right, this way. What have you got to say about this, Henry? I didn't do it, Dad. That's the truth? Yes, sir. It's good enough for me. Through here.  The shoes that your boy is wearing here. You got them for him? That's right. I had them made of water. Where? Place back east and main. You know of anyone else's a pair of them? No, as far as I know, that's the only pair on the coast. Good tennis shoes. I used to wear them when I played. Uh-huh. What do they have to do with this? Well, on each one of the steps, we found a print from a shoe like that outside the window. That doesn't mean that Alec was in them at the time. How about it, Alec? Do you let anybody else use the shoes? No. Well, there's your answer. Still doesn't come down to proof. What night did your boy spend out of the house, Mr. McCarron? I don't know. I guess he's gone on Friday most of the time. You know where he goes? The show, library. I don't keep a chain on him, Sergeant. I have complete trust in Alec. Burgers have been taking place on Friday nights? Still entirely circumstances. Yes, sir, but it fits, doesn't it? What's he supposed to have stolen? Costume jewelry. You mean five and dime stuff? Yes, sir. All right, there you are. You tell me why, sir? Why did Alec steal that? Now, look around, Sergeant. I just spent $1,500 on this room. Had it all redone. Alec's got just about everything he wants. He gets a good allowance. If he needs any more money, he only has to come to me. There's no reason for him to steal. I see this is a nice room and all. What you say may be true, but the fact stays, the thefts are taking place. All right, then I'll put it on the line. I've got faith in my son. He's had all the things a boy could want, a good family, good home, all the things that count. Maybe there is a thief, but it isn't my son. I know that. You got an explanation for the shoes for the Friday nights? No, I'll find one. Maybe we can try it for this, too, Mr. Karen, what? Found these in the boy's closet, hmm? A box of cheap costume jewelry. With the finding of the jewelry in Alec McCarron's room, the circumstantial case against him seems to solidify. Although he denied any knowledge of the pieces, he was taken down to George Street Juvenile and booked on suspicion of violation of section 459 P.C. to Lincoln. His father retained the lawyer immediately. The next day, the jewelry found at his home was shown to the victims, and they gave us a positive identification on each one of the pieces. A petition was filed and the facts were presented to the juvenile court. Detention was requested. On Thursday, July 8th, Frank and I got back from the second floor after talking to the McCarron boy. He still admitted nothing about the burglaries. I don't know, maybe I'm starting to believe the kid. It sure makes you wonder, doesn't it? Sergeant Frydy? Yes, sir. How are you, Mr. McCarron? Fine, thank you. I've got it, Sergeant. I've got it right here. Sir? The way to show you that my son had no part in these burglaries.  Now, you say that all the thefts have taken place on Friday night, right? Yes, sir. And your big piece of evidence is the heels on his shoes, right? Well, it's part of it, yes, sir. All right, then. You can just turn them loose. The big part? I remembered it last night. I couldn't get to sleep and got to thinking. I remembered. All right, go ahead. Alec's shoes get a lot of use. He plays a lot of tennis, wears his shoes all the time. Well, sir? The soles and heels wore out. We had to send them back east and have them resold. Sent them back on May 5th. Didn't get them until May 19th. Now, there's two Fridays in there, Sergeant. If the shoes were gone, Alec couldn't have done the burglaries. Right? Yeah. Well, I hope now that you admit that you've made a mistake. Well, if we have, Mr. McCarron, we'll sure tell you. Well, you can take my word for it. Alec had no part in any of this. He's good boys, honest, smart. He was going to steal anything. If it really came to that, he wouldn't risk his neck for a few cheap beads. Doesn't seem so, does it? Here you are, John. All right. Well, what about it? Why don't you find out? Just a minute, sir. Are you going to release my son now? No, sir. Well, the shoes were out of state. Then he couldn't have committed the thefts. Mm-hmm. Month of May? Two Fridays in that month. That's right. No burglaries, only the one of them. We're listening to Dragnet, the authentic story of your police force in action. It seemed like every attempt the father made to prove the innocence of Alec McCarron only turned up another piece of evidence against him. The date for his hearing in juvenile court was set, and Frank and I talked with Mr. Bowman in the probation department on the preparation of the case. From the evidence, it appeared that the boy would be found guilty. We questioned and requested him. We went over every facet of his story, but he still stuck to it. At no time did he change any part of it. Saturday, July 10th. Frank and I checked into the office. Boy, it's gonna be another scorching. Yeah, probably overnighty again. Yeah, overnighty. More than that, Joe. I'm going over the Broadway at noon today to pick up a pair of those shoes I was telling you about. Mm-hmm. With a tan suit. Ones with a hose in it. Not to suit the shoes, you know. Finally decided last night I need them more in the new spin-in reel. I'm gonna get them. Well, that's good. Anything for us? Yeah. Looks like we might be wrong about that McCarron boy. What do you mean? Another house-breaking last night. Footprints all over the place. Yeah. All of them with the same heel print. We left the office. We went over to the crime land. Ray Pinker showed us photographs that had been made at the scene. The marks were identical with those we'd found before. The size of the shoe, the way they were worn, everything matched. The information blew the case against Alec McCarron's sky high and both Frank and I felt a little relieved. The probation department was informed what had happened and the boy was released immediately. Frank and I drove him to his home where he was reunited with his family. We obtained the name and address of the shoe company who had made the tennis shoes and the radiogram was sent to the police department in that city requesting all information on similar shipments to the Los Angeles area. We got the answer the next morning. Only one other pair had been sent to the Southland. Those had been sent to a Mr. Homer Ellsworth, care of general delivery, Ventura, California. We checked the name through records and identification, but we were unable to come up with a record form. The name was sent to George Brick and CII Sacramento. At a quarter past 11, Frank and I drove up to the beach city. We talked to the postmaster and several of the clerks. They told us that several pieces of mail had been delivered to Ellsworth. Although none of the people could tell us where the man lived, they were able to give us a good description of him. They went on to say that they had gathered from conversations that he resided someplace in Los Angeles. We left word with them that the local police department be called and the event the man returned and then Frank and I drove back to the city. We checked the name again this time with a description. No name. One of the outstanding things in the way the man was described was that the little finger from his left hand was missing. This was checked through the oddity file without result. The additional information was forwarded to Brayerton. We checked with burglary detail, but they had nothing on the suspect. 8.40 p.m. We drove out to the McCarran house and talked again to his family. I knew it all the time. Got a boy like Alec after 16 years, you know, pretty much what he will, what he won't do. Couple things we'd like to go over with you. Sure, anything you want. Just let me know. Caught the person who was really responsible for this? That's what we want to talk to you about. Now, the description we've got of the man is that he's about 50, around 5 foot 8, 150 pounds, dark hair, gray at the temples, wears glasses, has a mustache. Got a finger missing from the left hand. What's his name? Well, we don't know, sir. That's why we're asking you. No name at all. Well, he's used Homer Ellsworth. We don't think that's his true name. It isn't, sir. Name's not Ellsworth, it's Fallon. Homer Fallon. You know him? Sure, he worked for me for 10 years. Sort of a handyman does our jobs around the house. You know where we can reach him? Not sure about that. What do you mean? He quit yesterday morning. We got Fallon's address. We left for the place right away. It was a small house in the south Hollywood area. The front yard was overgrown and several of the pickets from the fence were missing. It was a 1938 Chevrolet truck parked in the driveway. Frank and I left our unit and walked up to the front door. Watch your steps, Joe. Front warden lives there. Yeah, I see. All right, everyone. I got it. Don't hear anything. We must be here. I've been taking the truck with me. Yeah. Who are you? Police officers who want to talk to you. Come on in. Sit down. Just fixing a little dinner. Yes, sir, go right ahead. Spaghetti. Want some? No, sir. Pretty good. That kind of it's all cooked. Just drop it in the water and heat up the sauce. Pretty good. Sure you'd want some? No, thank you. How about you? No, sir. Well, wait a minute. I'll turn down the fire and we can talk. All right, sir. You've got to watch it pretty close. It's too cooked and you've got to eat it with a spoon. Not good that way. No, sir. You've had it, have you? You want to come in here, sir? Yeah, I'll be right with you. Open up the sauce and I'll be right there. Maybe you better come in now. What? Come on in. Sit down. You haven't got no right to tell me what to do in my own house. No right at all. Here, you've got no right to go through my things. We'll come back with a warrant if that'd be easier for you. And you go ahead and do it. You've got no right. Is there something you know on me to find? No. I've got nothing to hide. Then you won't mind. I will, too. You guys come in here while I'm having dinner and causing a lot of trouble. You've got no right to do this. I'm an old man. I can't take a lot of excitement. And why don't you save us all a lot of time? You know why we're here. You know what we want. I do? Sure you do. Tell us. Big thing over nothing. That's all it is. Then you tell us about it, will you? All this stuff put together wasn't worth a dollar and a half. None of it. Not even a dollar and a half. Then why'd you do it? I wanted to get the kid in trouble. All his life, he's had what he wanted. All his life, different from me. I never had nothing. You put the things we found in his room, did you? Yeah. I knew he'd get to it and wanted to sew it up, make it look good. I put him there. Why'd you do the burglary last night? Well, I got the feeling sorry for the kid sitting in jail. You ever been arrested, fella? Well, a long time ago. Where? Back east. Bag charge. I was just a kid working my way out here. Took me off a freight. I sat in the can overnight and then they let me go. Give me a floater out of town. I got sorry for the kid. I figured if there was another stealing while he was in jail you'd let him go. All right. Get your coat, will you? Yeah. Where's it? Closet there. The only one in there. I got it. You know, I was sorry right after. I didn't want to hurt the kid. He's really not too bad. Didn't want to cause him no trouble. You did pretty good, didn't you? I suppose so. Want to go? Yeah. Spaghetti. Got to turn it off. All right. I guess it doesn't make any difference anyway. What was that, sir? Don't make any difference about you taking me to jail. Is that right? Yeah. And I cooked the spaghetti too long. Merlin's e-pallin was tried and convicted of burglary in the first degree, six counts, and received punishment as prescribed by law. Burglary in the first degree is punishable by imprisonment for a period of less than five years in the state penitentiary. Dragnet is a presentation of the United States Armed Forces radio service.


I'm Second Lieutenant Nicholas Rivera, I'm the platoon commander of medium girder bridge platoon at Bridge Company 9PSP. I'm a 1302's command engineer officer. In 9th Engineer Support Battalion, we have a standard bridging company which builds the medium girder bridge and the improved ribbon bridge. I'm Staff Sergeant Green, I'm the platoon sergeant for MGB Platoon Bridge Company 9PSP in 1371. We provide mobility to the MLG and the MAGTAP as a whole, enabling wheeled track vehicles and personnel in order to get across various types of gaps. A non-standard bridge is any bridge that is not either the medium girder bridge, the standard bridges for the Marine Corps and the IRB. In the other category we have non-standard bridging, which is taking the readily available materials, whether it's timber, concrete, something as basic as trees and dirt, sandbags, hasco. Whatever it is that we have available, we identify as needed for that gap. We can employ that to cross wheeled track vehicles, or in this case, to mobile personnel. A unit having the capability to build a non-standard bridge gives them a lot of leeway where the standard bridges for the Marine Corps either don't fit the capabilities or their mission requirements of what we need to do.