Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to prove that if I omit the fact that $\mu (X) < \infty$ in Egorov theorem and place instead that our functions $|f_n| <g$ and $g$ is integrable, we still get the result of Egorov's theorem.

Fix $m$ a natural number.

I took $ w_{n} = |f_n-f|$ and thus by DCT $\int |f_{n} - f|$ goes to zero.

Then I took $\bigcup_n {( w_{n} \geq 1/m)}$. I need its measure to be finite. Its measure is less than the sum of the measures of each $ w_n\geq 1/m)$ varying $n$, and by Tchebychev, this is less than $m\int|f_{n} - f|$. But I got stuck here.

Any help is appreciated. Thanks!

share|cite|improve this question
up vote 10 down vote accepted
+50

We can, without loss of generality, assume that the sequence $f_n$ is decreasing and convergent almost everywhere to $0$ (if it's not the case, consider $\displaystyle g_n(x):=\sup_{k\geq n}|f_n-f(x)|$, which is dominated by $2g$, an integrable function. We fix $\varepsilon>0$, and we are looking for a measurable set $A$ such that $\mu(X\setminus A)\leq \varepsilon$ and the sequence $\{f_n\}$ converges uniformly to $f$ on $A$. We have for all integers $j$ and $n$ that $$\mu\left(\left\{g_n\geq \frac 1j\right\}\right)\leq j\int_X |g_n|d\mu,$$ and the monotone convergence theorem gives us that $\displaystyle\lim_{n\to\infty}\int_X |g_n|d\mu=0$. Hence for all $j\geq 1$, we can pick $n_j\in\mathbb N$ such that $$\mu\left(\left\{g_{n_j}\geq \frac 1j\right\}\right)\leq \varepsilon2^{-j}.$$ Now, put $A_j:=\left\{g_{n_j}\geq \frac 1j\right\}$ and $\displaystyle A:=\bigcap_{j\geq 1}\complement_X A_j$. We have $$\mu(X\setminus A)=\mu\left(\bigcup_{j\geq 1}A_j\right)\leq \sum_{j\geq 1}\mu(A_j)=\varepsilon\left(\frac 1{1-2^{-1}}-1\right)=\varepsilon$$ and $\displaystyle\sup_{x\in A}\,|g_{n_j}-0|\leq \frac 1j$. Since the sequence $\displaystyle\left\{\sup_{x\in A}\,g_n(x)\right\}$ is decreasing and has a sub-sequence which converges to $0$, the whole sequence converges to $0$, and we are done.

share|cite|improve this answer
    
What does $\complement_X$ mean? – Sujaan Kunalan Apr 12 '15 at 17:14
1  
@SujaanKunalan The complement in $X$. – Davide Giraudo Apr 12 '15 at 17:32
    
The complement of $A$ in $X$? – Sujaan Kunalan Apr 12 '15 at 17:34
    
Yes. ${}{}{}{} $ – Davide Giraudo Apr 12 '15 at 17:35

Here is another solution:

We can write:

$$X=\{ g>1\}\cup(\bigcup\limits_{k=1}^{\infty}\left\{ 2^{-k}<g\leqslant2^{1-k}\right\})\cup \{g=0\}$$ Notice that the above sets are all disjoint.

Apply Egoroff's theorem to each of the sets: $$\{ g>1\},\bigcup\limits_{k=1}^{\infty}\left\{ 2^{-k}<g\leqslant2^{1-k}\right\}$$

Then there exist $A\subset G=\{g>1\}$ ,and $A_k\subset G_k=\left\{ 2^{-k}<g\leqslant2^{1-k}\right\}, $ such that: $$\mu (A)<\varepsilon/2 , \mu (A_k)<\varepsilon 4^{-k} $$ and $$f_n\longrightarrow f$$uniformly in $E^c=(G\setminus A)\cup(\bigcup\limits_{k=1}^{\infty}G_k\setminus A_k)\cup \{f_n=0,\forall n\}$, where the last set in the union is a superset of $\{g=0\}$.

Now, it suffices to prove that $\mu (E)<\varepsilon$.

Indeed, after some simple set calculations (using the fact that we decomposed X in disjoint sets) we can obtain that: $$E=X\setminus\left\{(G\setminus A)\cup(\bigcup\limits_{k=1}^{\infty}G_k\setminus A_k)\cup \{f_n=0,\forall n\}\right\}\subset$$ $$(X\setminus(G\setminus A))\cap(X\setminus(\bigcup\limits_{k=1}^{\infty}G_k\setminus A_k))\cap (X\setminus\{g=0\})\subset$$ $$A\cup(\bigcup\limits_{k=1}^{\infty} A_k)$$ Therefore, $$\mu (E)\leqslant \mu (A\cup(\bigcup\limits_{k=1}^{\infty} A_k))\leqslant \mu(A)+\mu (\bigcup\limits_{k=1}^{\infty} A_k))<\varepsilon/2+\sum\limits_{k=1}^\infty \varepsilon4^{-k} =\varepsilon/2+\varepsilon/3 <\varepsilon$$

share|cite|improve this answer
    
Can you explain the last subset relation? I have no idea how it follows. – takecare Mar 5 at 6:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.