Prove that $||f||_{\infty}$ is the smallest of all numbers of the form $\sup\{|g(x)|: x\in X\}$, where $f=g$ ($\mu$ almost everywhere).
In addition, if $f$ is a continuous function on the measure space $\mathbb{R}^n,\mathcal{L},\lambda$ (i.e., Lebesgue measure), prove that $||f||_{\infty}=\sup\{|f(x)|:x\in\mathbb{R}^n\}$
Here's the book's definition: $L^{\infty}(X,\mathcal{M},\mu)$ is the collection of all essentially bounded $\mathcal{M}$-measurable functions on $X$. Moreover, the norm of a function $f\in L^{\infty}$ is the number $$||f||_{\infty}=\inf\{M:|f(x)|\le M\quad \text{for } \mu \text{-almost everywhere } x\in X\}$$
I think the statement in the second paragraph (i.e., assuming continuity of $f$ on the measure space) is a consequence of the first paragraph. If $f=g$ almost everywhere, and $f$ is continuous, does it follow that $f=g$ identically?
Best Answer
First part: Suppose $\lVert f\rVert_\infty = K$. Then for any $M > K$ we know $|f(x)| \le M$ almost everywhere. Choose $g$ to equal $f$ everywhere where this holds, and to equal 0 everywhere else. Then clearly $$\inf_{g} \sup{|g(x)|} \le K$$ I'll leave the converse to you.
Second part: No, that doesn't follow unless $g$ is continuous too. Instead, notice that if $|f(x)| \le M$ almost everywhere, then in particular in any open interval around $x_0$ there are points with $|f(x)| \le M$, and therefore a sequence of points converging to $x_0$ with $|f(x)| \le M$. You're done by continuity. You can do the other direction.