Exercise 11.1

Question

\def\where{\,|\,}
Exercise 1: If $f\ge 0$ and $\int_Ef\,d\mu=0$, prove that $f(x)=0$ almost everywhere on $E$.

analambanomenos · Accepted Answer

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Following the hint, let $E_n$ be the set of all $x\in E$ at which $f(x)>n^{-1}$, $n=1,2,\ldots,$ and let $K_{E_n}$ be the characteristic function of $E_n$. Then $0\le n^{-1}K_{E_n}<f$ on $E$, and so
    by Remark 11.23(c)
    $$
        0=\int_Ef\,d\mu>\frac{1}{n}\int_EK_{E_n}\,d\mu=\frac{1}{n}\,\mu(E_n)\ge 0
    $$
    so that $\mu(E_n)=0$ for all $n$. If $f(x)>0$, then $x\in E_n$ for some positive integer $n$, so $A=\cup E_n$ is the set at which $f(x)>0$. Since $\mu(A)=\lim\mu(E_n)=0$ by Theorem 11.3, $f(x)=0$
    almost everywhere on $E$.

Amit Awasthi · Answer

\begin{proof}
      Let $E_n = \{ x \in E \mid f(x) > 1/n \}$, and write $A = \bigcup E_n$.
      We want to show $\mu(A)= 0$. Now $\mu(A) = 0$ if and only if
      $\mu(E_n) = 0$ for all $n$: the forward direction is clear since $E_n
      \subset A$, and by monotonicity $(8)$, and the converse is Theorem
      $11.8(b)$. Now $f(x) = 0$ if and only if $\mu(A) = 0$, which is equivalent
      to $\mu(E_n) = 0$ for all $n$. But this is equivalent to having
      $\int_{E_n} f\,d\mu = 0$ for all $n$, which is in turn equivalent to
      having $\int_E f \,d\mu = 0$.
    \end{proof}

Exercise 11.1

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Exercise 11.1

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