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I want to prove the following:

Let $(\Omega, \mathcal A, \mu) $ be a measure space. Let $f$ be a measurable function and let there be a sequence of measurable sets $E_n \uparrow \{f\neq 0 \}$ with $\mu(E_n) < \infty$ for each $n\in \mathbb N$. Prove:
$$\lVert f\rVert_\infty = \sup \left \{ \int_\Omega \lvert fg\rvert \, d\mu : g\in \mathcal L^1(\Omega, \mathcal A, \mu), \lVert g \rVert_1 \leq 1 \right \}.$$

My attempt:
If $g\in \mathcal L^1(\Omega, \mathcal A, \mu)$ and $\lVert g \rVert _1 \leq 1$ then $$\int \lvert fg\rvert \, d\mu \leq \lVert f\rVert_\infty \rVert g \lVert _1 \leq \lVert f\rVert_\infty$$ by the Hölder inequality. Taking the supremum over all such $g$ gives us the $\geq$ part.

Now, for $\leq$, I am only able to prove it under the assumption that $\mu$ is $\sigma$-finite:
Let $\epsilon > 0$. Let $E_n \uparrow \Omega$ with $\mu (E_n) < \infty$ for each $n\in \mathbb N$. Define $N_\epsilon := \{ f > \lVert f \rVert _\infty - \epsilon\}$. Now we have $$N_\epsilon = N_\epsilon \cap \bigcup_{n\in \mathbb N}E_n = \bigcup_{n\in \mathbb N} (N_\epsilon \cap E_n)$$ and by continuity of $\mu$ there must be some $n\in \mathbb N$ with $\mu (N_\epsilon \cap E_n) > 0$. Call the corresponding set $M$. Fix such an $n\in \mathbb N$ and define $$g:= \frac{1_M}{\mu(M)}.$$ Clearly, $\lVert g \rVert_1 = 1$ and $$\int \lvert fg \rvert \, d\mu \geq \lVert f\rVert _\infty - \epsilon.$$ Letting $\epsilon \to 0$ proves the statement.

But how do I prove it under the weaker assumption in the exercise? Thanks for your help!

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Let $S=\{f\ne 0\}$ and $N_{\epsilon}=\{x\in S:|f(x)|>\|f\|_{L^{\infty}}-\epsilon\}$, then $\mu(N_{\epsilon})>0$ and $N_{\epsilon}=\displaystyle\bigcup_{n}(N_{\epsilon}\cap E_{n})$ and put $g=\dfrac{\chi_{M}}{\mu(M)}$ for $M=N_{\epsilon}\cap E_{n_{0}}$, where $\mu(N_{\epsilon}\cap E_{n_{0}})>0$.

Now $\displaystyle\int_{\Omega}|fg|d\mu=\int_{S}|fg|d\mu=\int_{M}\dfrac{|f|}{\mu(M)}d\mu\geq\|f\|_{L^{\infty}}-\epsilon$.

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  • $\begingroup$ Why is $\mu (N_\epsilon) > 0$? $\endgroup$ – Staki42 Feb 11 '18 at 21:06
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    $\begingroup$ Assume that $\mu(N_{\epsilon})=0$, then take an $x\in X-N_{\epsilon}$. If $x\in S^{c}$, then $|f(x)|=0\leq\|f\|_{L^{\infty}}-\epsilon$. If $x\in S$, then $|f(x)|\leq\|f\|_{L^{\infty}}-\epsilon$ by the construction of $N_{\epsilon}$, in either case, we have $|f(x)|\leq\|f\|_{L^{\infty}}-\epsilon$. But $N_{\epsilon}$ is of measure zero, then $\|f\|_{L^{\infty}}\leq\|f\|_{L^{\infty}}-\epsilon$, a contradiction. $\endgroup$ – user284331 Feb 11 '18 at 21:08
  • $\begingroup$ Oh I see. Thanks! $\endgroup$ – Staki42 Feb 11 '18 at 21:10

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