Exercise 11.2

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Exercise 2: If $\int_Af\,d\mu=0$ for every measurable subset $A$ of a measurable set $E$, then $f(x)=0$ almost everywhere on $E$.

analambanomenos · Accepted Answer

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Let $A^+$ be the measurable set where $f(x)\ge 0$ on $E$. Then $\int_{A^+}f\,d\mu=0$ implies that $f(x)=0$ almost everywhere on $A^+$ by Exercise 1. Similarly, if $A^-$ is the measurable set where
    $f(x)\le 0$ on E, then $\int_{A^-}(-f)\,d\mu=0$ implies that $f(x)=0$ almost everywhere on $A^-$. Since $E=A^+\cup A^-$, $f(x)=0$ almost everywhere on $E$.

Amit Awasthi · Answer

\begin{proof}
      Let $f(x) \ge 0$ on a measurable set $A$ and $f(x) \le 0$ on a measurable set
      $B$. Then, by Exercise \ref{11.1}, after restriction to $A$ and $B$,
      we see $f(x) = 0$ almost everywhere in $A$ and $B$, hence almost everywhere
      in $E$.
    \end{proof}

Exercise 11.2

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Exercise 11.2

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