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Background Information:

Theorem 1.18 - If $E\in M_{\mu}$, then \begin{align*} \mu(E) &= \inf\{\mu(U):E\subset U, U \ \text{open}\}\\ &=\sup\{\mu(K):E\subset K, K \ \text{compact}\}\end{align*}

Theorem 2.26 - If $f\in L^1(\mu)$ and $\epsilon > 0$, then

a.) there is an integrable simple function $\phi = \sum_{1}^{n}a_j\chi_{E_j}$ such that $\int |f - \phi|d\mu < \epsilon$.

b.) If $\mu$ is a Lebesgue-Stieltjes measure on $\mathbb{R}$, the sets $E_j$ in the definition of $\phi$ can be taken to be finite unions of open intervals.

c.) Moreover, in situation b.), there is a continuous function $g$ that vanished outside a bounded interval such that $\int |f - g|d\mu < \epsilon$.

Corollary 2.32 - If $f_n\rightarrow f$ in $L^1$, there is a sub-sequence $\{f_{n_j}\}$ such that $f_{n_{j}}\rightarrow f$ a.e.

2.33 Egoroff's Theorem - Suppose that $\mu(X) < \infty$, and $f_1,f_2,\ldots$ and $f$ are measurable complex-valued functions on $X$ such that $f_n\rightarrow f$ a.e. Then for every $\epsilon > 0$ there exists a set $E\subset X$ such that $\mu(E) < \epsilon$ and $f_n\rightarrow f$ and $f_n\rightarrow f$ uniformly on $E^c$.

Question:

Lusin's Theorem - If $f:[a,b]\rightarrow \mathbb{C}$ is Lebesgue measurable and $\epsilon > 0$, there is a compact set $E\subset [a,b]$ such that $\mu(E^c) < \epsilon$ and $f|E$ is continuous.

Attempted proof - Let $f:[a,b]\rightarrow\mathbb{C}$ be Lebesgue measurable and $\epsilon > 0$. By theorem 2.26 we can build a sequence of continuous functions $\{g_n\}$ such that $$g_n\rightarrow f \ \text{in} \ L^1$$ Then by Corollary 2.32 there is a sub-sequence $\{f_{n_j}\}$ of $\{g_n\}$ such that $f_{n_j}\rightarrow f$ a.e. Now, by Egoroff's theorem there exists a set $E$ with $\mu(E) < \infty$ such that $g_n\rightarrow f$ uniformly on $X\setminus E$. Note since $\{f_{n_j}\}$ is a sub-sequence of $\{g_n\}$ then we have $f_{n_j}\rightarrow f$ uniformly on $X\setminus E$ as well. Now by theorem 1.18 we can take $E$ to be a compact subset of $[a,b]$ since $K$ is compact and $E\subset K$. Note that $X\setminus E$ is the space on which $g_n\rightarrow f$ uniformly so then $E^c$ must be part of that space where $g_n\rightarrow f$ uniformly and thus $\mu(E^c) < \epsilon$ by definition. Finally since $\{g_n\}$ are continuous functions and $g_n\rightarrow f$ uniformly on $X\setminus E$ then $f$ is continuous.

I am not sure this is exactly correct, any suggestions is greatly appreciated.

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  • $\begingroup$ I think your proof is more or less correct, but you seem to confuse $E$ with $E^c$ toward the end. (It should read "and thus $\mu(E) < \epsilon$", I think.) The source of this confusion is probably the statements of Egoroff's and Lusin's theorems. In Egoroff, $E$ is the set with small measure, and in Lusin, it is $E^c$. It might be clearer if you restate one of them using $F$ instead of $E$, or interchange the roles of $E$ and $E^c$ in one of the theorems. $\endgroup$ – Bungo Jul 8 '16 at 2:18
  • $\begingroup$ Also, I think corollary 2.32 only gives you $f_{n_j} \to f$ almost everywhere, not everywhere, right? (This is all you need.) $\endgroup$ – Bungo Jul 8 '16 at 2:24
  • $\begingroup$ Ah ha your right it is almost everywhere sorry about that $\endgroup$ – Wolfy Jul 8 '16 at 2:39
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@Wolfy , you proof is essentially correct, but it has some confusion in terms of notation. Also it need some improvement inthe very last part. I have copied you proof, making the necessary adjustments.

Lusin's Theorem - If $f:[a,b]\rightarrow \mathbb{C}$ is Lebesgue measurable and $\epsilon > 0$, there is a compact set $E\subset [a,b]$ such that $\mu(E^c) < \epsilon$ and $f|E$ is continuous.

Proof - Let $f:[a,b]\rightarrow\mathbb{C}$ be Lebesgue measurable and $\epsilon > 0$.

Since $f$ is finite and $\mu([a,b])<\infty$, there is $K>0$ such that $\mu(\{x \in [a,b] : |f(x)|>K \})< {\epsilon}/{4}$. Let $$ H= \{x \in [a,b] : |f(x)|>K \}$$ we have that $\mu(H)< {\epsilon}/{4}$ and $f\chi_{H^c}: [a,b] \rightarrow \mathbb{C}$ is in $L^1[a,b]$.

By theorem 2.26 we can build a sequence of continuous functions $\{g_n\}$ such that $$g_n\rightarrow f\chi_{H^c} \ \text{in} \ L^1$$

Then by Corollary 2.32 there is a sub-sequence $\{g_{n_j}\}$ of $\{g_n\}$ such that $g_{n_j}\rightarrow f\chi_{H^c}$ a.e.. Now, by Egoroff's theorem, for any $\epsilon >0$, there exists a set $G\subset [a,b]$, with $\mu(G) < \epsilon/4$ such that $g_{n_j}\rightarrow f\chi_{H^c}$ uniformly on $G^c$.

(Atention: we can apply to Egoroff's theorem to the subsequence $g_{n_j}$, because we have that $g_{n_j}\rightarrow f\chi_{H^c}$ a.e.).

Let $F = H \cup G$. We have then $$\mu(F) \leqslant \mu(H) +\mu(G)= \epsilon/4 + \epsilon/4= \epsilon/2$$ and, since $F^c= H^c \cap G^c$, we have that $g_{n_j}\rightarrow f$ uniformly on $F^c$.

Now by theorem 1.18, since $\mu([a,b])<\infty$, there is $E$ a compact subset of $[a,b]$, such that $E\subset F^c$ and $$\mu(F^c)-\epsilon/2 < \mu(E)\leq \mu(F^c)$$
So $F \subset E^c$ and we have \begin{align*}\mu(E^c) &=\mu(F)+\mu( E^c \setminus F)= \\&= \mu(F)+\mu( E^c \cap F^c)= \\& = \mu(F)+\mu( F^c \setminus E)=\\&=\mu(F)+(\mu( F^c)- \mu(E)) \leq \\ & \leq \frac{\epsilon}{2}+\frac{\epsilon}{2} =\epsilon \end{align*} Note that, since $E\subset F^c$ and $g_{n_j}\rightarrow f$ uniformly on $F^c$, we have that $g_{n_j}\rightarrow f$ uniformly on $E$. Since, for all $j$, $g_{n_j}$ is continuous, we have that $f$ is continuous on $E$, that is $f|_E$ is continuous.

Remark: Here is a detailed proof that "Since $f$ is finite and $\mu([a,b])<\infty$, there is $K>0$ such that $\mu(\{x \in [a,b] : |f(x)|>K \})< {\epsilon}/{4}$"

Proof: For each $K\in\mathbb{N}$, let $A_K=\{x \in [a,b] : |f(x)|>K \}$. So $\{A_K\}_{K\in\mathbb{N}}$ is a non-increasing sequence of mensurable set. Since $f$ is finite, $\bigcap_{K\in\mathbb{N}}A_K=\emptyset$. Then, since $\mu([a,b])<\infty$, we have that

$$\lim_{K \to \infty}\mu(A_K) = \mu \left (\bigcap_{K\in\mathbb{N}}A_K\right)=\mu(\emptyset)=0$$

So, given any $\epsilon>0$, there is $K>0$ such that $\mu(\{x \in [a,b] : |f(x)|>K \})< {\epsilon}/{4}$

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  • $\begingroup$ Don't you need to adjust the argument slightly in the case where $f\notin L^1[a,b]$? $\endgroup$ – user3281410 Dec 8 '16 at 23:06
  • $\begingroup$ @user3281410 Yes. Thanks for pointing it. I have updated my answer. $\endgroup$ – Ramiro Dec 9 '16 at 15:37
  • $\begingroup$ Hi, perhaps you can answer this. I know this question is old. Can you justify when you say "Since f is finite..." and the fact you can make the set where f blows up of arbitrary small measure? I dont see how this is done as we are only given that f is measurable. In particular, I dont know how to assert that f is in $L^1$ so it is safe to invoke density of continuous function in $L^1$. $\endgroup$ – Nicholas Roberts Sep 28 '18 at 3:57
  • $\begingroup$ @NicholasRoberts 1. I have added a remark to explain in more detail that "Since $f$ is finite and $\mu([a,b])<\infty$, there is $K>0$ such that $\mu(\{x \in [a,b] : |f(x)|>K \})< {\epsilon}/{4}$". 2. I did not assert that $f$ is in $L^1[a,b]$. What I asserted is that $f\chi_{H^c}: [a,b] \rightarrow \mathbb{C}$ is in $L^1[a,b]$. Note that $ H^c= \{x \in [a,b] : |f(x)|\le K \}$ and so $f\chi_{H^c}$ is a bounded measurable function defined in a space of finite measure ($[a,b]$), so $f\chi_{H^c}$ is in $L^1$. $\endgroup$ – Ramiro Sep 30 '18 at 7:29
  • $\begingroup$ I appreciate that. But where in the question does it assume that f is finite? $\endgroup$ – Nicholas Roberts Sep 30 '18 at 15:03

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