Recall that $\|f\|_\infty = \inf\{M\ |\ |f(x)|\leq M \text{ for almost every } x \in X\}$, for every function $f$ defined on $X$. Show that, if $\|f\|_\infty$ is finite, then $\|f\|_\infty$ is also the smallest of all numbers of the form $\sup\{|g(x)| : x \in X \}$, where $f = g$ almost everywhere with respect to the Lesbegue measure.
I am trying to understand how $L^{\infty}$ is defined. Could someone help me see how to prove the above fact?
Let's rewrite the definition of $\lVert f\rVert_{\infty}$. The condition $\lvert f(x)\rvert \leqslant M$ a.e. can also be stated as
$$\mu(\{ x \in X : \lvert f(x)\rvert > M\}) = 0,$$
where $\mu$ is the measure under consideration. Thus, with
$$A(f) := \bigl\{ M \in [0,+\infty] : \mu(\{ x \in X : \lvert f(x)\rvert > M\}) = 0 \bigr\}$$
we have $\lVert f\rVert_{\infty} = \inf A(f)$. And in fact $A(f) = \bigl[\lVert f\rVert_{\infty}, +\infty]$.
Next, define
$$B(f) = \bigl\{ M \in [0,+\infty] : \bigl(\exists g\bigr)\bigl(g(x) = f(x) \text{ $\mu$-a.e., and } \sup_x \lvert g(x)\rvert \leqslant M\bigr)\bigr\}.$$
Then your task is to show that $A(f) = B(f)$ for every measurable function $f$.
Take $M \in A(f)$, and construct a $g$ that agrees almost everywhere with $f$ such that $\sup_x \lvert g(x)\rvert \leqslant M$. And take $M \in B(f)$ and deduce $\mu(\{ x\in X : \lvert f(x)\rvert > M\}) = 0$.