Suspicious corollary of Lusin’s Theorem

Question

Suppose $f : [0,1] \to \mathbb R$ is integrable. According to Lusin’s Theorem (as I understand it), for every $\epsilon > 0$, there is a closed subset $C \subset [0,1]$ for which $f|_C : C \to \mathbb R$ is continuous and $\lambda([0,1] \setminus C) < \epsilon$, where $\lambda$ is the Lebesgue measure on $[0,1]$. Furthermore, if $B \subset [0,1]$ is open, then there is a closed $C’ \subset B$ for which $f_{C’} : C’ \to \mathbb R$ is continuous and $\lambda(B \setminus C’) < \epsilon$.

Playing with this, I arrived at the following corollary, which seems way too powerful to be true:

Corollary: There is a subset $C \subset [0,1]$ of full measure, for which $f|_C : C \to \mathbb R$ is continuous (in the induced subspace topology on $C$).

Proof. Let $C_1 \subset [0,1]$ be a closed set for which $f|_{C_1}$ is continuous, with $\lambda([0,1] \setminus C_1) < 2^{-1}$, so $\lambda(C_1) > 2^{-1}$. The existence of such a $C_1$ follows from Lusin’s Theorem. Now let $C_2 \subset [0,1] \setminus C_1$ be such that $f|_{C_2}$ is also continuous, and $\lambda\left(\left([0,1]\setminus C_1\right)\setminus C_2\right) < 2^{-2}$, or $\lambda(C_1 \sqcup C_2) > 2^{-2}$. The existence of $C_2$ also follows from Lusin’s theorem. Continuing in this way, we get a countable collection of disjoint closed sets $(C_n)_{n \geq 1}$ for which $f|_{C_n} : C_n \to \mathbb R$ is continuous for each $n$, and for which $\lambda\left(C_1 \sqcup C_2 \sqcup \cdots \sqcup C_n\right) > 1-2^{-n}$. By lower semicontinuity of $\lambda$, denoting $C = \bigsqcup_{n\geq 1} C_n$, we have $$ 1 \geq \lambda(C) = \lim_{n \to \infty} \lambda\left(\mathop{\bigsqcup}_{k=1}^n C_k\right) \geq \lim_{n \to \infty}(1-2^{-n}) = 1. $$ So $\lambda(C) = 1$. and since $C$ is the countable disjoint union of closed sets, on each of which $f$ is continuous in the subspace topology, it follows that $f$ is continuous on $C$ in its subspace topology. QED.

I don’t see an obvious flaw with this proof, but I find this hard to believe. The only place I can see an issue possibly is with the existence of $C_n$ for $n \geq 2$ using Lusin’s Theorem, but Lusin’s Theorem should hold on any Radon measure space $(X, \mathcal A, \mu)$—in particular it should hold on any measurable subset of $[0,1]$. Where is the flaw in this proof? Or is this really a corollary of Lusin’s Theorem?

EDIT: See the answer below.

This doesn't answer your specific question of where the flaw is, but you may be interested in the mini-essay on Lusin's theorem I give in my answer to Is every Lebesgue measurable function on $\mathbb{R}$ the pointwise limit of continuous functions? — Dave L. Renfro, Jan 07 '21 at 19:18

score 4 · Answer 1 · answered Jan 07 '21 at 19:13

The flaw is in assuming $f|_C$ is continuous. As a friend pointed out to me, a function that is continuous on an exhaustive sequence of closed sets need not be continuous: you’d need to show that given $x \in C$ and $\epsilon>0$, we have $|f(x) - f(y)| < \epsilon$ for all $y$ sufficiently close to $x$ in the set $C$. But if we have $x \in C_n$, then we could have $|x-y|<\delta$ for small $\delta>0$ but $y \in C_m$, $m \neq n$. Since there are infinitely many $C_m$, this can happen infinitely often.

Suspicious corollary of Lusin’s Theorem

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