Show that if $f(x) > 0$ for all $x \in [a,b]$, then $\int_{a}^b f(x) dx > 0$

Question

Assume $f$ is Riemann integrable and nonnegative over $[a,b]$. Show that if $f(x) > 0$ for all $x \in [a,b]$, then $\int_{a}^b f(x) dx > 0$.

This seems very obvious to me. One thing I would do is if the function is not continuous, break up the integrals it is continuous on into infinitely many small integrals which all must a approach a positive number. I still am trying to see how to make this argument rigorous.

Is this for a Riemann integral? Or Lebesgue? Or something third altogether? — hmakholm left over Monica, Apr 26 '16 at 23:36
Why does the question say $f(x)$ is nonnegative then say $f(x) > 0$? — user19405892, Apr 27 '16 at 00:09
Duplicate: http://math.stackexchange.com/questions/228951/prove-int-ab-fx-mathrm-dx0-if-fx0-at-forall-x-in-a-b-and-riem — Kibble, Apr 27 '16 at 00:12
@user19405892: Could you please say what tools are at your disposal? Particularly, do you know that if $[a, b]$ is a countable union of measurable sets, then at least one set must have positive measure? — Andrew D. Hwang, Apr 27 '16 at 00:15
@AndrewD.Hwang This is a calculus question, so I don't think you need any of that. — user19405892, Apr 27 '16 at 00:16
Do you mean 1. the question arose (was assigned in) in a calculus course, or 2. the problem statement refers only to calculus, so you expect to find a proof using nothing but calculus? — Andrew D. Hwang, Apr 27 '16 at 00:23
@AndrewD.Hwang The question was assigned in a calculus course. — user19405892, Apr 27 '16 at 00:31

Ali · Answer 1 · 2016-04-27T00:41:36.350

11

One can show without much difficulty that a Riemann integrable function on a bounded interval has at least a point of continuity. (There exists infinitely many of them actually)Let c be a point of continuity of $f(x)$ on the interval $ [a,b]$ . Note that since $f(c)>0$ there exists a neighborhood about $x=c$ such that $f(x) > \frac{f(c)}{2}$ on that neighbourhood. Thus we can easily conclude that the integral is strictly positive.

edited Apr 27 '16 at 00:41

answered Apr 27 '16 at 00:23

Ali

334

No you do not. Integral of $f$ on the whole interval is bigger than integral over that small neighbourhood above. Result follows trivially – Ali Apr 27 '16 at 00:27
I agree. This is so simple and basically exactly the same proof of the statement when $f$ is assumed continuous everywhere. I was thinking it had to be more involved here, but simplicity wins! – Winther Apr 27 '16 at 00:40
The essence of the proof lies in the fact that f should be continuous at some point in the interval. This of course I have not written out as I am short on time but the idea is to use some sort of nested compactness argument. I can write it out later – Ali Apr 27 '16 at 00:46
This is a fairly well-known theorem proven in almost every book on this subject, you can probably just reference Wiki or e.q. this question. – Winther Apr 27 '16 at 00:48
A function is Riemann-integrable if and only if it is bounded and continuous except on a set of Lebesgue measure zero; hence the claim that there exists at least one point of continuity is in fact a drastic understatement. – Chill2Macht Apr 27 '16 at 00:50
Obviously; I was responding to those criticizing Ali's statement, to the effect that they are not only wrong, they are drastically so. – Chill2Macht Apr 27 '16 at 00:55
@Winther How does $f(x) > \dfrac{f(c)}{2}$ on an interval imply the integral is strictly positive? – user19405892 Apr 27 '16 at 16:36
@user19405892 Draw a sketch to see this. If $f(x) > C > 0$ on $[c,d]$ what can you say about the area below $f(x)$? What is the smallest area that is under the curve? The important thing here is that $f(x)$ is positive so the integral over the whole region $[a,b]$ is larger or equal than the integral over any small interval $[c,d]$. If you can show that this latter integral is positive then so is the whole integral. – Winther Apr 27 '16 at 16:42

score 0 · Answer 2 · answered Apr 27 '16 at 00:14

Hmm, going through the Lebesgue integral actually seems to be the simplest path here.

$f$, being Riemann integrable on a bounded interval, is actually Lebesgue integrable, and its Riemann integral equals its Lebesgue integral.

Now the preimages $$f^{-1}\bigl[(1/n,\infty)\bigr]$$ must have measure zero for every $n$, because if one of them has positive measure $m$, then the integral would be at least $\frac mn$, that is, positive.

But the union of these (countably many) preimages is all of $[a,b]$ if $f$ is positive everywhere; on the other hand a countable union of sets of measure zero will itself have measure zero. This contradicts the (unspoken) assumption that the interval $[a,b]$ is nontrivial.

score 0 · Answer 3 · answered Apr 27 '16 at 01:45

To show there is at least one point of continuity: At every $x$ we can consider $\omega (f)(x),$ the oscillation of $f$ at $x.$ Recall that $\omega (f)(x)=0$ iff $f$ is continuous at $x.$ So all we need is one $x$ where $\omega (f)(x)=0.$ If that fails, then $[a,b] = \cup_n \{\omega (f)(x) \ge 1/n\}.$ But it's not hard to see each of the sets in the union is closed. By Baire, then, one of these sets contains an interval $[a',b']\subset [a,b].$ But $f$ is Riemann integrable on $[a',b'].$ But on this interval,$ U(f,P) - L(f,P)\ge (b'-a')/n$ for every partition $P.$ That's a contradiction, giving us a point of continuity as desired.

Show that if $f(x) > 0$ for all $x \in [a,b]$, then $\int_{a}^b f(x) dx > 0$

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