The Theorems of Lusin and Egorov

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There are two important theorems which make statements about convergence types of measurable maps. They are important to understand the behavior of sequence of measurable maps and to understand the importance of uniform convergence.

Theorem (Egorov)

Let [math](E,\A,\mu)[/math] be a measure space. Let [math]f_k:E\longrightarrow \bar \R[/math] be measurable for all [math]k\in\mathbb{N}[/math] and [math]f:E\longrightarrow\bar \R[/math] be measurable and [math]\mu[/math]-a.e. finite. Moreover [math]f_k(x)\xrightarrow{k\to\infty} f(x)[/math] [math]\mu[/math]-a.e. for [math]x\in E[/math]. Then for all [math]\delta \gt 0[/math] there exists [math]F\subset E[/math], with [math]F[/math] compact and [math]\mu(E\setminus F) \lt \delta[/math] and

[[math]] \sup_{x\in F}\vert f_k(x)-f(x)\vert\xrightarrow{k\to\infty} 0, [[/math]]

i.e. [math](f_k)_{k\in\mathbb{N}}[/math] converges uniformly to [math]f[/math] in [math]F[/math].

Show Proof

Let [math]\delta \gt 0[/math]. For [math]i,j\in\mathbb{N}[/math] set

[[math]] C_{i,j}:=\bigcup_{k=j}^{\infty}\{x\in E\mid \vert f_k(x)-f(x)\vert \gt 2^{-1}\}. [[/math]]

[math]C_{i,j}[/math] is [math]\mu[/math]-measurable, because [math]f[/math] and [math]f_k[/math] are [math]\mu[/math]-measurable and [math]C_{i,(j+1)}\subset C_{i,j}[/math], [math]\forall i,j[/math]. We also know that [math]f_k(x)\xrightarrow{k\to\infty}f(x)[/math] for [math]\mu[/math]-a.e. [math]x\in E[/math] and since [math]\mu(E) \lt \infty[/math] it follows that for all [math]i\in\N[/math]

[[math]] \lim_{j\to\infty}\mu(C_{i,j})=\mu\left(\bigcup_{j=1}^\infty C_{i,j}\right)=0. [[/math]]

So for every [math]i[/math] there exists a [math]N(i)\in\N[/math] with

[[math]] \mu(C_{i,N(i)}) \lt \delta\cdot 2^{-i-1}. [[/math]]

Now set [math]A=E\setminus \bigcup_{i=1}^\infty C_{i,N(i)}[/math]. Then

[[math]] \mu(E\setminus A)\leq \sum_{i=1}^{\infty}\mu(C_{i,N(i)}) \lt \delta/2, [[/math]]

and for all [math]i\in\N[/math] and [math]k\geq N(i)[/math]

[[math]] \sup_{x\in A}\vert f_k(x)-f(x)\vert \leq 2^{-i}. [[/math]]

Choose a [math]F\subset A[/math], where [math]F[/math] is compact with [math]\mu(A\setminus F) \lt \delta/2[/math]. Hence we have

[[math]] \mu(E\setminus F)\leq \mu(E\setminus A)+\mu(A\setminus F) \lt \delta. [[/math]]

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Theorem (Lusin)

Let [math](E,\A,\mu)[/math] be a measure space. Let [math]f:E\longrightarrow\bar \R[/math] be measurable and [math]\mu[/math]-a.e. finite. Then for all [math]\delta \gt 0[/math] there exists [math]F\subset E[/math], [math]F[/math] compact with [math]\mu(E\setminus F) \lt \delta[/math] and [math]f\mid_F:F\longrightarrow \R[/math] is continuous.

Show Proof

We split the proof onto two parts.

• We are going to show this theorem for step functions of the form
  [[math]] g=\sum_{i=1}^Ib_i\one_{B_i}, [[/math]]
  where we set [math]E=\bigsqcup_{i=1}^{I}B_i[/math] with [math]B_i\cap B_j=\varnothing[/math] for [math]i\not=j[/math]. For [math]\delta \gt 0[/math] choose [math]F_i\subset B_i[/math] compact with
  [[math]] \mu(B_i\setminus F_i) \lt \delta\cdot 2^{-i},1\leq i\leq I [[/math]]
  Since the sets [math]B_i[/math] are disjoint, it follows that the sets [math]F_i[/math] are also disjoint, because of the fact that they are also compact it follows that [math]d(F_i,F_j) \gt 0[/math] for [math]i\not=j[/math]. Therefore we notice that [math]g[/math] is locally constant, i.e. continuous on [math]F:=\bigcup_{i=1}^IF_i\subset E.[/math] Moreover [math]F\subset E[/math] and
  [[math]] \mu(E\setminus F)=\mu\left(\bigcup_{i=1}^I(B_i\setminus F_i)\right)\leq \sum_{i=1}^I\mu(B_i\setminus F_i) \lt \delta. [[/math]]
• Let [math]f_k:E\longrightarrow \R[/math] be a step function with
  [[math]] f(x)=\lim_{k\to\infty}f_k(x),x\in E, [[/math]]
  where
  [[math]] f_k=\sum_{j=1}^k\frac{1}{j}\one_{A_j}=\sum_{i=1}^{I_k}b_{ik}\one_{B_{ik}},k\in\N, [[/math]]
  with [math]B_{ik}\cap B_{jk}=\varnothing[/math] for [math]i\not=j[/math] and [math]\bigsqcup_{i=1}^{I_k}B_{ik}=E[/math] and with
  [[math]] b_{ik}=\sum_{B_{ik}\subset A_j}\frac{1}{j},1\leq i\leq I_k,k\in\N. [[/math]]
  For [math]\delta \gt 0[/math], [math]g=f_k[/math] choose compact sets [math]F_k\in E[/math] as in part [math](i)[/math] with
  [[math]] \mu(E\setminus F_k) \lt \delta\cdot 2^{-k-1},f_k|_{F_k}:F_k\longrightarrow \R\text{continuous,}k\in\N. [[/math]]
  Choose also [math]F_0\subset E[/math] compact with
  [[math]] \mu(E\setminus F_0) \lt \delta/2,\sup_{x\in F_0}\vert f_k(x)-f(x)\vert\xrightarrow{k\to\infty} 0. [[/math]]
  Finally let [math]F=\bigcap_{k=0}^\infty F_k\subset E[/math] Note that [math]F[/math] is compact with
  [[math]] \mu(E\setminus F)\leq \mu\left(\bigcup_{k=0}^\infty(E\setminus F_k)\right)\leq \sum_{k=0}^\infty\mu(E\setminus F_k) \lt \delta. [[/math]]
  and because of the fact that [math]F\subset F_0[/math] it follows that
  [[math]] \sup_{x\in F}\vert f_k(x)-f(x)\vert\xrightarrow{k\to\infty} 0. [[/math]]
  The continuity of [math]f_k\mid_{F}[/math], [math]k\in\N[/math], gives us now the continuity of [math]f\mid_F:F\longrightarrow \R.[/math]

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General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].