Graph of measurable function is measurable (and has measure zero)

Let f be a finite real valued measurable function on a measurable set E\subseteq\mathbb{R}. Show that the set \{(x,f(x)):x\in E\} is measurable.

We define \Gamma(f,E):=\{(x,f(x)):x\in E\}. This is popularly known as the graph of a function. Without loss of generality, we may assume that f is nonnegative. This is because we can write f=f^+ - f^-, where we split the function into two nonnegative parts.

The proof here can also be found in Wheedon’s Analysis book, Chapter 5.

The strategy for proving this question is to approximate the graph of the function with arbitrarily thin rectangular strips. Let \epsilon>0. Define E_k=\{x\in E\mid \epsilon k\leq f(x)<\epsilon (k+1)\}, k=0,1,2,\dots.

We have |\Gamma (f,E_k)|_e\leq\epsilon |E_k|, where |\cdot|_e indicates outer measure.

Also, \Gamma(f,E)=\cup\Gamma(f,E_k), where \Gamma(f,E_k) are disjoint.

\begin{aligned}|\Gamma(f,E)|_e&\leq\sum_{k=1}^\infty|\Gamma(f,E_k)|_e\\ &\leq\epsilon(\sum_{k=1}^\infty|E_k|)\\ &=\epsilon|E| \end{aligned}

If |E|<\infty, we can conclude |\Gamma(f,E)|_e=0 and thus \Gamma(f,E) is measurable (and has measure zero).

If |E|=\infty, we partition E into countable union of sets F_k each with finite measure. By the same analysis, each \Gamma(f,F_k) is measurable (and has measure zero). Thus \Gamma(f,E)=\bigcup_{k=1}^\infty\Gamma(f,F_k) is a countable union of measurable sets and thus is measurable (has measure zero).

Unknown's avatar

Author: mathtuition88

Math and Education Blog

2 thoughts on “Graph of measurable function is measurable (and has measure zero)”

Leave a comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.