Let $f: E \subseteq \mathbb{R}^n \to \mathbb{R}$ be a measurable function. The main idea behind Lusin's theorem is:

  • Construct a sequence of simple functions $\{f_k\}$ that converge to $f$
  • Each $f_k$ is nearly continuous (on $E \setminus E_k$ for some set $E_k$ of arbitrarily small measure)
  • By Egorov, $\{f_k\}$ is nearly uniformly convergent, and since each $f_k$ is nearly continuous, $f$ is continuous on the appropriate subset of $E$.

My confusion appears on point 2; namely, how can we say that each simple $f_k$ is nearly continous?


More formally, the claim is that we can find $E_k \subseteq E$ such that $m(E_k) < \epsilon$ for arbitrary $\epsilon$ where $f_k$ is continuous on $E \setminus E_k$. Why can we do this?

I know that since $f_k$ is simple, it takes a constant value on disjoint measurable sets (let's say $S_1, \ldots, S_m$) that partition $E$. Thus, $f_k$ should only be discontinuous on the boundaries $\partial S_1 \cup \ldots \cup \partial S_m$. But that union may not have arbitrarily small measure...

  • 2
    $\begingroup$ Point 2 is a nontrivial lemma that is nevertheless standard in measure theory and should appear in any textbook that contains Lusin's Theorem. $\endgroup$ May 10, 2017 at 0:42

1 Answer 1


Here's the main idea for point 2: Assume $m(E) < \infty.$ Let's look at your collection $\{S_1,\dots,S_m\}:$ Using the regularity of Lebesgue measure, we can say that inside each $S_j$ is a compact $K_j$ such that $m(S_j\setminus K_j)$ is as small as we like. Thus we will have

$$\sum_{j=1}^{m}c_j\chi_{S_j} = \sum_{j=1}^{m}c_j\chi_{K_j}$$

except on a set of small measure. Now because $K_1,\dots, K_m$ are pairwise disjoint compact sets, $\sum_{j=1}^{m}c_j\chi_{K_j}$ is continuous on $\cup_{j=1}^mK_j!$ Thus we've eliminated your worry about the boundaries of the sets $S_j$ by replacing them with the $K_j$'s, using inner regularity.

This is the main idea. Feel free to ask questions on this.

  • $\begingroup$ Ah, okay. So because $K_1, \ldots, K_m$ are compact and pairwise disjoint, we have that $d(K_i, K_j) > 0$ for all distinct $i, j$ giving us continuity. $\endgroup$ May 11, 2017 at 18:54
  • 1
    $\begingroup$ @LanguagesNamedAfterCofee Right, that positive distance gives us the breathing room we need. I just added an exclamation point to emphasize that. $\endgroup$
    – zhw.
    May 11, 2017 at 18:59

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