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In an exercise session of an analysis course (which covers Riemann integration and differentiation in one dimension rigorously) the following question came up:

Suppose $f$ is strictly positive and integrable (on some compact interval $[a,b]$ on the real line). Can we show that $\int_a^b f > 0$?

The proof by using measure theory and Lebesgue integration is easy, but also beyond the students at this point. So can one do without machinery such as sets of zero measure?

Tools in use: Mean value theorems, fundamental theorem of calculus, Riemann's condition for integrability (upper and lower integral within every epsilon of each other implies integrability), Riemann-Darboux integral, usual integration and differentiation techniques, as well as elementary real analysis in the epsilon-delta style.

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Hint: interval sequence theorem and upper Darboux sum –  89085731 Apr 13 '12 at 14:57
There's also a proof via the Baire category theorem. –  Chris Eagle Apr 13 '12 at 19:08

2 Answers 2

up vote 3 down vote accepted

This can be shown using the concept of oscillation.

For a bounded $f$, the oscillation of $f$ over set $A$ (which is not a single point) is given by

$$w(A) = \sup_{A} f - \inf_{A} f$$

For a single point $x$, the oscillation is defined as

$$ w(x) = \inf_{J} w(J)$$

where $J$ ranges over bounded intervals containing $x$.

Note that if $x \in I$, then $w(x) \le w(I)$.

Now if $f$ is Riemann integrable over $[a,b]$, then we can show that given any $n \gt 0$, there is a sub-interval $I_{n}$ of $[a,b]$ such that $w(I_n) \le \frac{1}{n}$.

This is because, if every subinterval $I$ of $[a,b]$ had $w(I) \gt \frac{1}{n}$, then for every partition of $[a,b]$ the difference between the upper and lower sums would be at least $\frac{b-a}{n}$ and as a consequence, $f$ would not be integrable.

Now pick $I_{n+1} \subset I_{n}$ such that $w(I_{n+1}) \le \frac{1}{n+1}$.

By completeness there is a point $c$ such that $c \in I_n$.

Thus $w(c) \le w(I_n) \le \frac{1}{n}$ for all $n$. Hence $w(c) = 0$.

Now it can be show that $f$ is continuous at point $x$ if and only if $w(x) = 0$.

Note: This is basically a simplification of one proof of the Riemann Lebesgue theorem of continuity almost everywhere.

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I think it may not be the Riemann Lebesgue theorem.The Riemann Lebesgue Theorem might be $\lim_{n\to \infty}\int_a^b f(x)g(nx)dx=\frac{1}{T}\int_a^b f(x)dx\int_0^T g(x)dx$ –  89085731 Apr 14 '12 at 0:26
@Gingerjin: There are multiple with "Riemann Lebesgue" name. For instance, there is the Riemann Lebesgue Lemma... –  Aryabhata Apr 14 '12 at 0:29

If f is integrable in [a,b], there is a point x0 in the interval where f is continuous, and positive. Then there is a neighborhood of the point where f is positive , contained in the interval.There we can take a closed inreval where it is positive and there f has minimum then the integral at this interval is bigger or equal than the integral of the minimum which is positive. We can then apply aditivity in the interval [a,b] and in the other closed intervals the integral is no negative since each riemann summ si no negative.

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I think you first need to prove there is a point that is continuous in[a,b] for him.Actually, this conclusion will lead to the conclusion that the continuous point is dense. –  89085731 Apr 13 '12 at 15:05
Yes.We can use that if f is Riemann integrable then the set of discontinuities has measure zero, but I will think how to prove without measure theory that there is a point of continuity. –  alpha.Debi Apr 13 '12 at 15:56

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