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Let $f: [a,b]\to\mathbb R$ integrable. The following statements are equivalent:

$1) \int_a^b |f(x)|dx=0;$

$2) \mbox{ if } f \mbox{ is continuous at point } c \mbox{ then } f(c)=0;$

$3) \mbox{ int}(X=\{x\in[a,b]; f(x)\neq 0\})=\emptyset.$

For $1\Longrightarrow 2$ i tried:

Since $f$ is integrable, it follows that $|f|$ is integrable and $$ \left|\int_a^b f(x)dx\right|\leq\int_a^b|f(x)|dx=0 \Longrightarrow \int_a^b f(x)dx=0. $$

Moreover, since $f$ is continuous in $c$, then the function $$ F(x)=\int_a^x f(t)dt $$ is derivable in $c$ and $F'(c)=f(c)$.

Now, if $F'(c)\neq0$ then either $F'(c)>0$ or $F'(c)<0$. Assume $F'(c)>0$. Thus, there is $\delta>0$ such that $$ x,y\in(a,b) \mbox{ and } c-\delta<x<c<y<c+\delta \Longrightarrow F(x)<F(c)<F(y). $$

I'm stuck here. Any tips?

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    $\begingroup$ I'm guessing from your first step that the integral in question is the Lebesgue integral? If that is the case, I suggest tagging this question with lebesgue-integral $\endgroup$ Nov 2, 2018 at 18:25
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    $\begingroup$ @TheoBendit Is there a reason it can't be Riemann integration? $\endgroup$
    – Arthur
    Nov 2, 2018 at 18:27
  • $\begingroup$ it is riemann integrable $\endgroup$
    – nom
    Nov 2, 2018 at 18:28
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    $\begingroup$ @Arthur I guess not. I confused $f$ being integrable $\implies$ $|f|$ is integrable with its converse. $\endgroup$ Nov 2, 2018 at 18:30

1 Answer 1

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Hint:

Work only with $|f|$. The positivity of $\int_a^b f(x) \,dx$ does not guarantee that $f(c) = 0$.

Since $||f(x)| - |f(c)|| \leqslant |f(x) - f(c)|$, the function $|f|$ is continuous at $c$. If $|f(c)| > 0$ then $|f(x)| > |f(c)|/2 > 0$ on some interval $I \subset [a,b]$. Now find a contradiction.

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