Lusin's Theorem. Let $f$ be a real-valued measurable function on $E$. Then for each $\varepsilon>0$, there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F$ contained in $E$ for which $f=g$ on $F$ and $m(E\sim F)<\varepsilon$.

Let us suppose we have proved the case when the measure of $E$ is finite.

$m(E)= +\infty $ Case Proof Attempt. We can write $E$ as $E=\bigcup_{n=1}^{\infty}E_n$ where $\{E_n\}_{n=1}^{\infty}$ is a collection of disjoint measurable sets of finite measure. So for each $n \in \mathbb{N}$, $m(E_n)<\infty$ and $f$ is measurable on $E_n$. Then by the finite case, for each $n \in \mathbb{N}$ and $\varepsilon>0$, there exist a continuous function $g_n$ on $\mathbb{R}$ and a closed set $F_n \subseteq E_n$ such that on $F_n$: $$f=g_n,$$ and $$m(E_n \sim F_n)<\frac{\varepsilon}{2^{n}}.$$

Define $F:=\bigcap_{n=1}^{\infty}F_n$. Choose any $N \in \mathbb{N}$. Then we have a closed set $F$ such that $F \subseteq F_n \subseteq E_n \subseteq E$, and a function $g:=g_N$ continuous on $\mathbb{R}$ such that on $F \subseteq F_N$: $$f=g,$$ and $$m(E \sim F) = m \left(\bigcup_{n=1}^{\infty}( E_n \sim F_n) \right) \leq \sum_{n=1}^{\infty} m(E_n \sim F_n)< \sum_{n=1}^{\infty} \frac{\varepsilon}{2^{n}} = \varepsilon,$$ as required.

Can anybody check if my proof is valid?

  • $\begingroup$ You should say at the beginning that $E\subset \mathbb R.$ $\endgroup$
    – zhw.
    Mar 19, 2016 at 17:11
  • $\begingroup$ Try $F:=\bigcup_{n=1}^\infty F_n$. In this case, you need to work harder to prove that $F$ is closed. $\endgroup$
    – user0
    Mar 24, 2021 at 19:34

1 Answer 1


It is not valid.

Since the $E_n$ are disjoint, so are the $F_n$, and therefore $F = \emptyset$. $$E \sim F \ne \bigcup_{n=1}^{\infty}( E_n \sim F_n)$$


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