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I read a good number of posts which deal with this specific topic, but none of them seem to answer my question. If this is a repost, I apologize.

Alright so here is the problem, followed by a theorem which I use, and then my attempted proof:

Exercise 7.4.4. Show that if $f(x)>0$ for all $x\in [a,b]$ and $f$ is integrable, then $\int_{a}^{b} f>0$.

The theorem (and exercise) is from Abbott's Understanding Analysis, and my proof uses Theorem 7.4.2(ii):

Theorem 7.4.2.(ii) Assume $f$ is integrable on $[a,b]$. If $m\leq f(x)\leq M$ for all $x\in [a,b]$ then $$m(b-a)\leq\int_a^b f(x) \leq M(b-a)$$

Now here is my proof:

Proof. Let $f(x)>0$ for all $x\in [a,b]$ and assume $f$ is integrable. Since $f$ is integrable, then we can let (but can we?) $$m=\inf\{f(x_0):x_0 \in [a,b]\}\text{ and }M=\sup\{f(x_0):x_0\in [a,b]\}$$ Where both $m$ and $M$ are greater than zero since $f(x)>0$. Now by Theorem 7.4.2.(ii), we can write $$m(b-a)\leq \int_{a}^{b} f\leq M(b-a)$$ And since $0<m(b-a)\leq M(b-a)$, then the integral must be greater than zero as well.
Q.E.D.

I would just like some verification/correction. My doubt is whether or not the values $m$ and $M$ necessarily have to exist, since Theorem 7.4.2.(ii) states that the inequality is only true IF there exists such values for $m$ and $M$. But since $f$ is integrable and $[a,b]$ is compact, then the function should attain a maximum and a minimum, correct?

JB071098
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  • You have "strictly increasing" in the title and then that disappears below? – zhw. Apr 11 '18 at 03:19
  • I misspoke. I meant to have "Strictly Positive", which I have fixed. @zhw. – JB071098 Apr 11 '18 at 03:23
  • Use the fact that $f$ is continuous at some point in $[a, b] $ (see https://math.stackexchange.com/a/519921/72031) and hence positive in a subinterval so that the integral is positive. – Paramanand Singh Apr 11 '18 at 08:20

1 Answers1

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That is not correct. Since Riemann integrability implies boundedness, both infimum and supremum exist, but if we let $f(x)=x$ for $x\in(0,1]$ and $f(0)=1$, unfortunately, $\inf_{x\in[0,1]}f(x)=0$ and the infimum is not attainable.

Elementary Way?

Assume the contrary that $\displaystyle\int_{a}^{b}f(x)dx=0$, then find a partition $P=\{x_{0},...,x_{n}\}$ such that $U(f,P)\leq b-a$. It cannot be the case that all $\sup_{x\in[x_{i-1},x_{i}]}f(x)>1$, $i=1,...,n$, so some $i$ is such that $\sup_{x\in[x_{i-1},x_{i}]}f(x)\leq 1$, let us write $\sup_{x\in[a_{1},b_{1}]}f(x)\leq 1$.

Now $\displaystyle\int_{a_{1}}^{b_{1}}f(x)dx=0$. Choose another partition $Q$ in $[a_{1},b_{1}]$ such that $U(f,Q)\leq\dfrac{b_{1}-a_{1}}{2}$, then by the same reasoning we have some subinterval $[a_{2},b_{2}]$ of $[a_{1},b_{1}]$ such that $\sup_{x\in[a_{2},b_{2}]}f(x)\leq\dfrac{1}{2}$.

Proceed in this way we have a decreasing chain $\{[a_{n},b_{n}]\}$ such that $\sup_{x\in[a_{n},b_{n}]}f(x)\leq\dfrac{1}{n}$.

Take $c\in\displaystyle\bigcap_{n=1}^{\infty}[a_{n},b_{n}]$, then $f(c)\leq\dfrac{1}{n}$ for all $n=1,2,...$, now taking $n\rightarrow\infty$, we have $f(c)=0$.

user284331
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    So is the only method of proving this with complete generality to use Lebesgue Integration/Measure Theory(measure/content zero)? The only posts regarding this topic use concepts from measure theory which I am not familiar with. – JB071098 Apr 11 '18 at 01:56
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    Let me work for a while whether any elementary method can come into my mind. – user284331 Apr 11 '18 at 01:59
  • You may take a look now. – user284331 Apr 11 '18 at 02:16
  • I see why this works, and I understand the nested interval property, but just to clarify, why can you assume that there always exists $\sup_{x\in[x_{i-1},x_i]}f(x)\leq 1$? If $f(x)=2$ for all $x\in [a,b]$ then the supremum is always $2$. – JB071098 Apr 11 '18 at 02:20
  • If all $\sup_{i}f>1$, then $U(f,P)=\displaystyle\sum_{i}(\sup_{i}f)\Delta x_{i}>\displaystyle\sum_{i}\Delta x_{i}=b-a$, contradicts the choice. – user284331 Apr 11 '18 at 02:21
  • So going back to my original argument, (with intention of using the given theorem) could I have just stated in a similar manner that there exists some $M>0$ such that $f(x)\leq M$ where $M$ satisfies the same conditions for our chosen $\epsilon = b-a$, and then proceeded with the same argument? – JB071098 Apr 11 '18 at 02:27
  • But then the $\epsilon$ is big, what can be concluded? And we know that $m$ could be zero. – user284331 Apr 11 '18 at 02:30
  • Actually I think your idea cannot proceed further, so if we let $\epsilon=b-a$, then this is fixed, and maybe you can conclude something, but then do we have the next step, especially can we shrink the suprema? To deduce that the integral is zero, we need some sort of infinite process, but I fail to see what is the appropriate next step. – user284331 Apr 11 '18 at 02:38
  • Well, since we are assuming that $f$ is integrable, then we can use the sequential criterion for integrals, i.e. we know that there is some sequence of Partitions $(P_n){n=1}^{\infty}$ satisfying $\int{a}^{b} f= \lim\limits_{n\rightarrow\infty}U(f,P_{n})$. Since we assumed the integral is zero, then for every $\epsilon$, we should be able to have that $\int_{a}^{b} f=\lim U(f,P_{n})=0<\epsilon$ Correct me if I am wrong. – JB071098 Apr 11 '18 at 02:48
  • This is almost the same idea that what I have written. – user284331 Apr 11 '18 at 02:48
  • Okay. I added a different version of what you have written using the Limit criterion for integrals. – JB071098 Apr 11 '18 at 03:00
  • That is not correct. We have $\lim U(f,p_{n})=0<\epsilon$ cannot deduce that $f(c)=0$ for some $c$, at least, not that immediately. – user284331 Apr 11 '18 at 03:01
  • I think your proof will suffice. Thank you for all of your help, as elementary as my questions may seem. – JB071098 Apr 11 '18 at 03:18