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Let $f:\Bbb R \to \Bbb R$ be continuous.

Does $f \geq 0 \forall x$ and $f>0$ for a countable set of points $A$ imply $\int_a^b f(x) dx >0$ if $A\subseteq (a,b)$?

I've had this doubt. I'm fairly sure this is not true, but I think it would take a pretty weird function to disprove it.

Does anyone know a simple proof/counterexample?

YoTengoUnLCD
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  • It is true. If $f$ is nonzero at even a single point then by continuity there is a neighborhood around that point where $f$ is nonzero. The integral over that neighborhood will necessarily be positive. – Alex G. Nov 14 '15 at 19:53

2 Answers2

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If $f(x)$ is continuous, $f(x)\geq 0$ and $f(x)>0$ for a SINGLE point, then $\int_a^b f(x) dx>0$

Use continuity to get a small interval around the SINGLE point $x_0$, where $f(x)>\frac{f(x_0)}{2}$ using the $\epsilon-\delta$ definition of a limit.

Edit, see for example $f\geq 0$, continuous and $\int_a^b f=0$ implies $f=0$ everywhere on $[a,b]$

Community
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It suffices that $f(x)>0$ for only one point : indeed, $U:=f^{-1}(]0,+\infty[)=\{x\in[a,b]\mid f(x)>0\}$ is an open set. If it is non-empty, then chose $x_0\in U$ and $r>0$ such that $B(x_0,r)\subset U.$ Then : $$0<\int_{x_0-r}^{x_0+r} f(x)\,dx\leq\int_a^bf(x)\,dx.$$

Balloon
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