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Prove that if $A \subset \Bbb R^n$ has a positive Jordan measure and $f$ is positive valued function on $A$ such that $\int_A f$ exists, then $\int_A f > 0.$

I am having trouble with the writing of this proof. I am trying to prove that the Riemann integration of $f$ on $A$ will be positive. How do I prove this for when $A$ is closed? I am trying to use $\forall \epsilon > 0$ $\text{vol}\{x \in A : f(x) \ge \epsilon \}= 0$ and from here maybe compactness could prove?

I search on this website for help but I could not find any, please let me know if this is repost.

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The idea is that $$ \int_A f \geq \mbox{vol}(A) \cdot \inf_{A} f > 0, $$ where the first inequality follows by standard properties of the Riemann integral and the latter inequality follows from positivity of $f$ and the positive Jordan measure of $A$. (Of course, we could still have $\inf_A f = 0$, so we have to be a bit more careful.)

I'll elaborate. Since $\int_A f$ exists, $f$ is continuous almost everywhere on $A$. Write $A$ as the disjoint union $A = C \cup D$, where $C$ is the subset of $A$ on which $f$ is continuous and $D$ is the subset of $A$ on which $f$ is discontinuous. Since $f$ is continuous almost everywhere on $A$, we have $\mbox{vol}(D) = 0$, and then $$ \int_A f = \int_C f + \int_D f = \int_C f, $$ where we have $\int_D f = 0$ because $\int_D f \leq \sup(f|_D) \cdot \mbox{vol}(D) = 0$. The virtue of this is that we may now work with $C$ instead of $A$; from the point of view of the integral, they're the same (this is what the last equation says), but we get the dividend that $f$ is continuous on $C$, where it wasn't necessarily on $A$.

Now, by definition, the Riemann integral is always bigger than a lower Darboux sum, so pick a finite collection $P$ of mutually disjoint boxes contained in $C$, yielding the lower Darboux sum $L(f,P)$: $$ L(f, P) = \sum_{B \in P} \inf (f|_B) \cdot \mbox{vol}(B) \leq \int_C f. $$ Now, $$ L(f,P) = \sum_{B \in P} \inf (f|_B) \cdot \mbox{vol}(B) \geq K \sum_{B \in P} \mbox{vol}(B) > 0 $$ for some $K > 0$. In the last step, $K > 0$ exists because $f$ is continuous and positive on a compact set (a box $B$), so it attains a minimum value (say $K_B$) on $B$ which must be positive. (Then take $K = \min_{B \in P} K_B$, which is also positive.)

Combining everything tells us that $\int_A f = \int_C f \geq L(f,P) > 0$.

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  • $\begingroup$ Thank you, your answer is very complete $\endgroup$ – Jerry Crowers Dec 8 '15 at 5:45
  • $\begingroup$ One question,why does $\int_A' f = \int_A f$? I understand when you say it is continuous almost everywhere but I am not exactly understanding why you can conclude that it is continuous everywhere. $\endgroup$ – Jerry Crowers Dec 8 '15 at 6:05
  • $\begingroup$ I'll edit my answer to clarify this; let me know if it's insufficient. $\endgroup$ – Jon Warneke Dec 8 '15 at 16:49
  • $\begingroup$ Ok I fully understand now, thank you! $\endgroup$ – Jerry Crowers Dec 8 '15 at 18:01

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