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The question (Abbott, Understanding Analysis 2ed, 7.4.4) is:

Show that if $f(x)>0$ for all $x\in[a,b]$ and $f$ is integrable, then $\int_a^b f>0$.

I can show it using Baire's theorem (the sets $E_n=\{x: f(x)>1/n\}$ can't all be nowhere dense...), but that's optional in this book, and the Lebesgue characterization of integrable functions is two sections ahead. Is there a way using not much more than the definition of Riemann integral?

dubya
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    Maybe using limit of Riemann sums or properties of upper-lower integrals? – AspiringMathematician Apr 18 '17 at 03:19
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    If we are using Riemann integral, we are assuming $f$ is Riemann integrable, no? So $f$ must be continuous at a point $a$ and thus is at least $\frac{a}{2}$ in an interval around it. If we are using Lebesgue integration then one of $E_n$ has positive measure (no need for Baire). – Reinstate Monica Apr 18 '17 at 03:20
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    @Solomonoff'sSecret That is good enough for an answer. :) –  Apr 18 '17 at 03:22
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    You should be able to adapt the answer to this question, at least to show that $f$ is continuous at one point (and then the result is easy), but possibly into a more direct proof. – Chappers Apr 18 '17 at 19:42
  • @Chappers: Thanks, clearly my searching needs some work. It seems that the best answer is to at least prove enough of the Lebesgue characterization of Riemann integrable functions to get continuity at one point. I don't think it's a good problem for that section of the book. – dubya Apr 24 '17 at 14:46

2 Answers2

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The statement is true because one can show there exists $\xi \in (a,b)$ such that $$\int_a^b f(x) \, dx \geq f(\xi) (b-a)$$

In fact assume that $$f(x) > \frac 1{b-a} \int_a^b f(y)\,dy$$ for all $x \in (a,b)$

Changing the values of $f$ at $a$ and $b$ if necessary (which doesn't alter the value of the integral), we have a new function $\hat{f}$ such that $$\hat{f}(x) > \frac 1{b-a} \int_a^b \hat{f}(y)\,dy$$ for all $x \in [a,b]$

Remember now the theorem discussed in this question.

Since $\hat{f}$ is continuous at some $c \in (a,b)$, we can find $\varepsilon > 0\,$ so that $$\hat{f}(x) > \frac 1{b-a} \int_a^b \hat{f}(y)\,dy + \varepsilon$$ for all $x \in (c-\varepsilon,c+ \varepsilon) \subset (a,b)$.

Then, if we consider the partition $P=\{a,c-\varepsilon,c+\varepsilon,b\}$, we obtain $$\int_a^b \hat{f}(x)\,dx \geq L(\hat{f},P) \geq \int_a^b \hat{f}(x)\,dx + 2\varepsilon^2 > \int_a^b \hat{f}(x)\,dx$$ which is absurd.

The paper Rodrigo Lopez Pouso, Mean Value Integral Inequalities , Real Anal. Exchange Volume 37, Number 2, (2011), 439-450 is worth reading not only as the source of this proof.

Tony Piccolo
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    Hmm, if we know there is a point of continuity $c$ this could be a bit simpler. Choose $\epsilon = f(c) / 2$; then there is a $\delta$ for which $f(x) > \epsilon$ for $x \in [c - \delta, c + \delta]$. Then $\int_a^b f(x)dx \geq 2\delta \epsilon$. – Jair Taylor May 05 '17 at 20:07
  • @JairTaylor Surely but my aim was drawing attention to that mean value inequality. – Tony Piccolo May 05 '17 at 21:35
  • Also that paper on mean value inequalities is a great read. – Paramanand Singh May 06 '17 at 04:47
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As a sketch, try to write a Riemann-type sum with terms $f(x^*_j) \Delta x_j$ and notice that all terms in the sum are positive. Go by comparison.

Sean Roberson
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    I believe this approach shows $\lim_{|P| \to 0} \sum f(x_j) \delta x_j = \int_a^b f \geqslant 0 $ but not $\int_a^b f > 0$. If every term of a sequence is positive you can't rule out the possibility the limit is $0$, eg. $1/n > 0$ but $1/n \to 0$. – RRL May 06 '17 at 06:47
  • Don't we have $\int_{a}^b f = \text{sup } L(P,f) = \text{sup} \sum \text{min}f(x_j) \Delta x_j \geq \sum \text{min} f(x_k) \Delta x_k > 0$ since all the terms in the sum are positive if $f$ is positive? Here, $L(P,f)$ denotes the lower riemann sum of $f$ with respect to a partition $P$. Am I making a mistake somewhere? – A. B. Marnie May 08 '17 at 17:09
  • Oh, nevermind, I see that my definition of $L(P,f)$ is incorrect. We really have $L(P,f) = \sum \text{inf} f(x_k) \Delta x_k$. The infimum of $f$ may be zero, even when $f > 0$. – A. B. Marnie May 08 '17 at 17:24