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I'm very confused about $L^\infty$. So I'm trying to prove this:

Is $\|f\|_{\infty}$ the smallest of all numbers of the form $\sup\{|g(x)| \,:\,x \in X\}$, where $f=g $ $\mu$-a.e.?

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up vote 9 down vote accepted

The answer to your question is yes. Recall that the definition of $\|f\|_\infty$ is $$\|f\|_\infty=\operatorname{ess sup}(f)=\inf\{a\in\mathbb{R}\mid \mu(\{x\in X\mid |f(x)|>a\})=0\}.$$ I'll interpret the phrase "the smallest of all numbers of the form $\sup\{|g(x)|\,:\,x\in X\}$, where $f=g$ $\mu$-a.e." to mean $$M=\inf_{g=f\text{ a.e.}}\left\{\sup_{x\in X}|g(x)|\right\}.$$

For any $\epsilon>0$, we can define $g:X\to\mathbb{R}$ by $$g(x)=\begin{cases}f(x) & \text{if }|f(x)|\leq \|f\|_\infty+\epsilon,\\0 & \text{otherwise}.\end{cases}$$ Note that $$\mu(\{x\in X\mid f(x)\neq g(x)\})=\mu(\{x\in X\mid |f(x)|>\|f\|_\infty+\epsilon\})=0$$ by the definition of $\|f\|_\infty$, so that $f=g$ almost everywhere, and that $\sup_{x\in X}|g(x)|\leq \|f\|_\infty+\epsilon$.

Thus, $$M=\inf_{g=f\text{ a.e.}}\left\{\sup_{x\in X}|g(x)|\right\}\leq \|f\|_\infty+\epsilon$$ for all $\epsilon>0$, and thus $M\leq \|f\|_\infty$.

For the other direction, note that for any $\epsilon>0$, $$\mu(\{x\in X\mid |f(x)|>(\|f\|_\infty-\epsilon)\})>0$$ by the definition of $\|f\|_\infty$. If $g=f$ a.e., then certainly $|g|=|f|$ a.e., so we must have that $$\mu(\{x\in X\mid |g(x)|>(\|f\|_\infty-\epsilon)\})>0$$ too, so that $g(x)>\|f\|_\infty-\epsilon$ for some $x\in X$. Thus, for any $g$ such that $g=f$ a.e., we have that $$\sup_{x\in X}|g(x)|\geq \|f\|_\infty-\epsilon$$ so that $$M=\inf_{g=f\text{ a.e.}}\left\{\sup_{x\in X}|g(x)|\right\}\geq \|f\|_\infty-\epsilon$$ for all $\epsilon>0$, and thus $M\geq \|f\|_\infty$.

This shows that $M=\|f\|_\infty$.

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It's perfect answer. – Collins Jun 13 '12 at 6:33
Thanks, glad to help! – Zev Chonoles Jun 13 '12 at 6:34
Just a little note: $\operatorname{ess sup}(f)$ should be $\operatorname{ess sup}(|f|)$. – matheburg Aug 7 '14 at 11:25

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