Take the 2-minute tour ×

Suppose you have 2 probability spaces $(\Omega, \mu)$ and $(\Psi,\lambda)$. For every $t\in (0,1)$ let $f_t$ be a real-valued non-negative measurable function bounded by one, that is $f_t: \Psi\times\Omega \rightarrow [0,1]$, such that one has $$ \frac{1}{\mu(A)\lambda(B)}\int_{A \times B} f_t(x,w) ~d\mu(w) d\lambda(x) \xrightarrow[]{t\rightarrow 0} f~~~ $$ $$ \text{for every } A\times B \subset\Psi\times\Omega~\text{with nonzero measure}, $$ where $f$ is a fixed real number in $[0,1]$. The number $f$ does not depend on the choices of $A$ or $B$.

I would like to show that the function $x\mapsto \int_{\Omega} f_t (x,w)~d\mu(w)$ converges for $t\rightarrow 0$ almost surely to the constant function $f$. I have a strong feeling that this is true. However I don't know how to approach. If it makes things easier, I am especially interested in the case where $(\Omega,\mu) = (\Psi,\lambda) = ([0,1], \text{Lebesgue measure})$.

share|improve this question
The condition $f$ does not depend on the choices of "A" and "B" looks strange. Did you mean to divide the integral by $\mu(A)\lambda(B)$? –  user31373 Jun 3 '12 at 23:39
If the sequence $(f_{t})$ is indexed by $t\in (0,1)$, how can you observe limiting when $t\to\infty$? –  Thomas E. Jun 4 '12 at 4:06
Sorry, of course I want to divide by $\mu(A)\lambda(B)$ and I am looking at $t\rightarrow 0$. –  h.h.543 Jun 4 '12 at 8:48

1 Answer 1

Here is a counterexample. Let $\phi_n$ be a sequence of functions $\phi_n\colon [0,1]\to[0,1]$ such that $\int_0^1\phi_n\to 0$ but $\phi_n$ does not converge a.e. (For example, take the characteristic functions of $[k/m,(k+1)/m]$, $0\le k<m$, $m\ge 1$, and arrange them into a sequence.) Define $f_t(x,w)=\phi_n(x)$ where $n=\lfloor 1/t\rfloor$.

For every product $A\times B\subset [0,1]^2$ with positive measure we have $\int_{A\times B}f_t = \lambda(B)\int_A \phi_n \to 0$. Yet, for any fixed $x$ the functions $\int f_t(x,w)\,dw = \phi_n(x)$ do not have a limit as $t\to 0$.

You can modify this example (adding a constant) to have nonzero constant $f$, if you wish.

share|improve this answer

Your Answer


