Egoroff‘s Theorem

Theorem

Assume $E$ has finite Lebesgue measure. Let { f n } \{f_n\} {fn​} be a sequence of Lebesgue measurable functions on $E$ that converges pointwise on $E$ to a Real-Valued function $f$. Then for each $ϵ>0$, there is a closed set $F$ contained in $E$ for which

{ f n } → f  uniformly on  F  and  m ( E − F ) < ϵ \{f_n\}\rightarrow f \text{ uniformly on } F \text{ and } m(E-F)<\epsilon {fn​}→f uniformly on F and m(E−F)<ϵ

proof

1. Under the assumptions of Egoroff’s Theorem, for each η>0 and δ>0, there is a Lebesgue measurable subset A of E and an index N for which |fₙ-f|<η on A for all n ≥ N and m(E-A)<δ.

2. By step 1, for each $n\in \mathbb{N}$, $ϵ>0$, let $\delta =ϵ/2^{n+1}$, $\eta =1/n$, there is a Lebesgue measurable subset $A_n$ of $E$ and an index $N(n)$ s.t.

   $|f_k-f|<\eta$

   on $A_n$ for all $k>N(n)$ and $m(E-A_n)<ϵ/2^{n+1}$

3. Define

   $$A=\bigcap _{n=1}^{\infty }{A}_n$$

4. By De Morgan’s Identities,the countably subadditivity of measure and $m(E-A_n)<ϵ/2^{n+1}$,

   $$m(E-A)=m\left(\bigcup _{n=1}^{\infty }[E-A_n]\right)⩽\sum _{n=1}^{\infty }m(E-A_n)⩽\sum _{n=1}^{\infty }ϵ/2^{n+1}=\frac{ϵ}{2}$$

5. We claim that { f n } \{f_n\} {fn​} converges to $f$ uniformly on $A$.

6. By step 2, $\forall ϵ>0$ there are $\frac{1}{n_0}<ϵ$, $A_{n_0}$ and $N(n_0)$ s.t.

   $$|f_k-f|<1/n_0$$

   on $A_{n_0}$ for all $k⩾N(n_0)$.

7. Since $A\subset A_{n_0}$ and step 6, { f n } \{f_n\} {fn​} converges to $f$ uniformly on $A$.

   $$\forall ϵ>0,\exists N\in \mathbb{N}(n>N\to |f_k-f|<ϵ)$$

   on $A$.

8. Since $A$ is Lebesgue measurable and inner approximation by closed sets and union of a countable collection of closed sets, we may choose a closed set $F\subset A$ for which $m(A-F)<ϵ/2$.

9. Thus $m(E-F)<ϵ$ and { f n } → f \{f_n\}\rightarrow f {fn​}→f uniformly on $F$.

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