Theorem 1.3.28 (Lusin's theorem I)

Question

Let $f:\mathbb{R}^d	o\mathbb{C}$ be an absolutely integrable function, and let $\epsilon>0$. Then there exists a Lebesgue measurable set $E \subseteq \mathbb{R}^d: m(E) \leq \epsilon$ such that $f|_{\mathbb{R}^d/E}$ is continuous on $\mathbb{R}^d/E$.

Elu Thingol · Accepted Answer

Since $f$ is absolutely integrable, by Theorem 1.3.20 (Approximation of $L^1(\mathbb{R}^d	o\mathbb{C})$ functions), for the given $\epsilon>0$ we can find a continuous and compactly supported functions $(f_n)_{n\in\mathbb{N}}$ such that
$$\| f - f_n \|_{L^1} = \int_{\mathbb{R}^d} |f-f_n| \leq \frac{\epsilon}{4^n}$$
By Markov's inequality (Lemma 1.3.15), we have for any $n\in\mathbb{N}$:
$$ \frac{1}{2^n} \cdot m \left( \left\{ x\in\mathbb{R}^d: |f(x) - f_n(x)| \geq \frac{1}{2^n}  ight\} ight) \leq \frac{\epsilon}{4^n}$$
which directly implies
$$\forall x \in \mathbb{R}^d/E_n: \quad | f_n(x) - f(x) | \leq \frac{1}{2^n}$$
for the Lebesgue measurable set $E_n = \left\{ x\in\mathbb{R}^d: |f(x) - f_n(x)| \geq {1}/{2^n}  ight\}$ with $m(E_n) \leq \epsilon/2^n$. Define
$$E:= \bigcup_{n=1}^{\infty} E_n$$
with $m(E) \leq \epsilon$. Then we have obtained uniform convergence outside of $E$ since
$$\forall n' \geq n: \quad d_\infty(f_{n'}|_{E^c},f|_{E^c})\leq \frac{1}{2^n} $$
But the uniform limit $f|_{E^c}$ of continuous functions $(f_n|_{E^c})_{n\in\mathbb{N}}$ must be again continuous (cf. Analysis II, Theorem 3.3.1). Since $m(E) \leq \epsilon$ we are done.

Theorem 1.3.28 (Lusin's theorem I)

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Theorem 1.3.28 (Lusin's theorem I)

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