This is the statement of Lusin's Theorem (taken from Royden):

Let $f$ be a real-valued measurable function on $E$. Then for each $\epsilon>0$, there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F\subseteq E$ for which $f=g$ on $F$ and $m(E\setminus F)<\epsilon$.

Q1) Taking $\epsilon=\frac 1k$, where $k\in\mathbb{N}$, we get a continuous function $g_k$ and a closed set $F_k\subseteq E$, for which $f=g_k$ on $F_k$ and $m(E\setminus F_k)<\frac 1k$.

What I am curious is, can we further assume that the $F_k$ are nested, i.e. $$F_1\subseteq F_2\subseteq F_3\subseteq\dots$$? This may be useful for some proofs. Intuitively, since $F_k$ are getting bigger, it seems possible but I am not sure how to prove it.

Update: I thought of replacing $G_k=\bigcup_{i=1}^k F_i$, but it doesn't work.



1 Answer 1


Yes, we can insure the desired nestedness. First: If $A,B$ are closed disjoint subsets of $ \mathbb R,$ then there exists a continuous $\varphi $ on $\mathbb R$ such that $\varphi = 1$ on $A,$ $\varphi = 0$ on $B.$

So in the problem at hand, there is a closed $F_1\subset E$ with $m(E\setminus F_1) < 1$ and a continuous $g_1$ such that $g_1=f$ on $F_1.$ We can then choose a closed $E_2\subset (E\setminus F_1)$ with $m((E\setminus F_1)\setminus E_2) < 1/2$ and a continuous $h_2$ such that $h_2=f$ on $E_2.$ Note that $(E\setminus F_1)\setminus E_2 = E\setminus (F_1\cup E_2).$ Now find a continuous $\varphi$ as above with $\varphi = 1$ on $F_1,$ $\varphi = 0$ on $E_2.$ The continuous function $g_2 =\varphi g_1 + (1-\varphi ) h_2$ then equals $f$ on $F_2 = F_1\cup E_2.$ This process can be continued to obtain the desired result.


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