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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.

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  • $\begingroup$ Is this for a Riemann integral? Or Lebesgue? Or something third altogether? $\endgroup$ Apr 26, 2016 at 23:36
  • $\begingroup$ It is a Riemann integral. $\endgroup$ Apr 26, 2016 at 23:38
  • $\begingroup$ Do you know the mean value theorem (for integrals)? $\endgroup$ Apr 26, 2016 at 23:41
  • $\begingroup$ @EricTowers Yes, I do. $\endgroup$ Apr 26, 2016 at 23:43
  • $\begingroup$ Why does the question say $f(x)$ is nonnegative then say $f(x) > 0$? $\endgroup$ Apr 27, 2016 at 0:09

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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.

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  • $\begingroup$ No you do not. Integral of $f$ on the whole interval is bigger than integral over that small neighbourhood above. Result follows trivially $\endgroup$
    – Ali
    Apr 27, 2016 at 0:27
  • $\begingroup$ 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! $\endgroup$
    – Winther
    Apr 27, 2016 at 0:40
  • $\begingroup$ 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 $\endgroup$
    – Ali
    Apr 27, 2016 at 0:46
  • $\begingroup$ 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. $\endgroup$
    – Winther
    Apr 27, 2016 at 0:48
  • $\begingroup$ 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. $\endgroup$ Apr 27, 2016 at 0:50
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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.

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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.

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