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Let $f$ be continuous and Riemann integrable on $[a, b]$ and $f(x) \geq 0$ for all $x \in [a,b]$.

I'm trying to show that if $\int^b_a f(x) \ dx = 0$ implies that $f(x) = 0$ for all $x \in [a,b]$.

Could someone give me a hint? I really do not know where to begin.

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Pick a point where $f$ is positive and use continuity to show it $f$ is positive nearby. – Brett Frankel Apr 18 '12 at 3:56
@BrettFrankel: Let $f(c) > 0$ for $c \in [a, b]$. By the definition of continuity, let $\epsilon$ be given. Then there is a $\delta$ such that $|x-c| < \delta \implies |f(x) - f(c)| < \epsilon$. – Student Apr 18 '12 at 4:00
Good. Now pick your $\varepsilon$ wisely. – Brett Frankel Apr 18 '12 at 4:01
@BrettFrankel: I don't see a good epsilon to pick in this case. Could you explain? – Student Apr 18 '12 at 4:06
@Jon Let $\epsilon$ be so small that $f(x)\ge f(c)/2$ for all $x$ in a $\delta$-nhood of $c$. – David Mitra Apr 18 '12 at 4:07
up vote 2 down vote accepted

Otherwise, there exists $x_0\in [a,b]$, such than $f(x_0)>0$. Without loss any generality , we may assmume that $x_0\in (a,b).$ Due to $\lim\limits_{x\to x_0}f(x)=f(x_0)>0$, there exists $\delta>0$, such that $(x_0-\delta,x_0+\delta)\subset(a,b)$ and $f(x)\geq\frac{1}{2}f(x_0).$ So $$\int_{a}^bf(x) dx\geq\int_{x_0-\delta}^{x_0+\delta}\frac{1}{2}f(x_0) dx=\delta f(x_0)>0.$$ And here comes the contradiction.

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I just have a question about: $\int_{x_0-\delta}^{x_0+\delta}\frac{1}{2}f(x_0) dx = \delta f(x_0)$. You can make this equality because $\int_{x_0-\delta}^{x_0+\delta}\frac{1}{2}f(x_0) dx = \int_{x_0}^{x_0+\delta} f(x_0) dx$ right? – Student Apr 18 '12 at 14:22
yes,youcan understand in this way. But I thiank the following is easy to understand :$\int_{x_0-\delta}^{x_0+\delta}\frac{1}{2}f(x_0)dx=\frac{1}{2}f(x_0)\int_{x_0-\‌​delta}^{x_0+\delta}dx= 2\delta\frac{1}{2}f(x_0)=\delta f(x_0).$ – Riemann Apr 18 '12 at 14:36

First,I want to say that the condition "$f$ is Riemann integrable on $[a,b]$ " can be deleted because "$f$ is continuous" implies "$f$ is Rieman integrable".

Prove by contradiction.If not,then $\exists x_0\in[a,b]$ such that $f(x_0)> 0$.$f$ is continuous on $[a,b]$,which means $\exists \varepsilon>0$ such that $\forall t\in (x_0-\varepsilon,x_0+\varepsilon)$,$|f(t)-f(x_0)|\leq \frac{f(x_0)}{2}$.So $\forall t\in (x_0-\varepsilon,x_0+\varepsilon)$,$\frac{f(x_0)}{2}\leq f(t)\leq \frac{3f(x_0)}{2}$.Now set a [partition] $P$ of $[a,b]$ such that $x_0-\varepsilon,x_0+\varepsilon\in P$.Then $L(f,P)\geq \varepsilon f(x_0)$(Why?).Because $\int_a^bf(x)dx\geq L(f,P)$(Why?),so $\int_a^bf(x)dx\geq \varepsilon f(x_0)>0$,this contradicts "$\int^b_a f(x) \ dx = 0$".So $\forall x\in [a,b]$,$f(x)=0$.

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