2014-11-04 16:52:11 8 Comments

How can I prove that if a sequence of functions $\{f_n\}$ that converges to $f$ in measure on a space of finite measure, then there exists a subsequence of $\{f_n\}$ that converges to $f$ almost everywhere?

### Related Questions

#### Sponsored Content

#### 0 Answered Questions

### Convergence in measure implies uniform convergence on a set of finite measure

**2016-05-07 05:43:06****user311526****512**View**2**Score**0**Answer- Tags: measure-theory convergence lebesgue-measure uniform-convergence

#### 1 Answered Questions

### [SOLVED] Convergence and Measure

**2018-12-04 21:56:43****ICanMakeYouHateME****46**View**0**Score**1**Answer- Tags: real-analysis measure-theory convergence lebesgue-measure

#### 1 Answered Questions

### [SOLVED] Convergence in $L^1$ implies there exists a subsequence that converges almost everywhere.

**2018-11-29 05:03:56****carsandpulsars****30**View**0**Score**1**Answer- Tags: real-analysis analysis measure-theory convergence measurable-functions

#### 1 Answered Questions

### [SOLVED] Convergence of integrals implies a.e. convergence for subsequence

**2017-10-23 15:37:43****Francesco Carzaniga****294**View**2**Score**1**Answer- Tags: measure-theory lebesgue-integral lebesgue-measure pointwise-convergence

#### 1 Answered Questions

### When does convergence in measure implies existence of subsequence converging almost everywhere

**2016-05-06 15:04:40****Anonymous****98**View**0**Score**1**Answer- Tags: real-analysis

#### 3 Answered Questions

### [SOLVED] Convergence in measure and almost everywhere

**2012-12-20 12:38:01****user53800****7070**View**7**Score**3**Answer- Tags: real-analysis measure-theory convergence

#### 1 Answered Questions

### [SOLVED] Convergence in measure implies convergence almost everywhere (on a countable set!)

**2014-12-14 17:17:17****Prism****1158**View**6**Score**1**Answer- Tags: real-analysis measure-theory convergence

#### 2 Answered Questions

### [SOLVED] When does almost everywhere convergence imply convergence in measure?

**2013-11-25 08:19:32****seriously divergent****6136**View**13**Score**2**Answer- Tags: measure-theory convergence

#### 1 Answered Questions

### [SOLVED] Converges in measure then a subsequence converges almost everywhere

**2013-05-09 03:18:39****Gaston Burrull****2403**View**2**Score**1**Answer- Tags: measure-theory proof-writing

#### 0 Answered Questions

### Lebesgue's dominated convergence theorem also holds replacing "almost everywhere convergence" by "convergence in measure"

**2012-10-31 20:05:51****LFRC****176**View**1**Score**0**Answer- Tags: measure-theory

## 1 comments

## @saz 2014-11-04 17:10:55

Let $(X,\mathcal{A},\mu)$ be a finite measure space and $(f_n)_{n \in \mathbb{N}}$ such that $f_n \to f$ in measure, i.e.

$$\mu(|f_n-f|>\varepsilon) \stackrel{n \to \infty}{\to} 0$$

for any $\varepsilon >0$. Setting $\varepsilon=2^{-k}$, $k \in \mathbb{N}$, we can choose $n_k$ such that

$$\mu(|f_n-f|> 2^{-k}) \leq 2^{-k}$$

for all $n \geq n_k$. Without loss of generality, $n_{k+1} \geq n_k$ for all $k \in \mathbb{N}$. Set

$$A_k := \{x \in X; |f_{n_k}(x)-f(x)| > 2^{-k}\}.$$

As $$\sum_{k \geq 1} \mu(A_k) \leq \sum_{k=1}^{\infty} 2^{-k} < \infty,$$ the Borel-Cantelli lemma yields

$$\mu \left( \limsup_{k \to \infty} A_k \right) =0.$$

It is not difficult to see that this is equivalent to

$$\lim_{k \to \infty} f_{n_k}(x) =f(x)$$

$\mu$-almost everywhere.

## @JonSK 2016-01-17 01:07:39

where did you use $(X,\mathcal{A},\mu)$ was a finite measure space?

## @saz 2016-01-17 07:33:31

@JonSK I used the Borel Cantelli lemma.

## @Matthew Kvalheim 2018-08-11 02:45:58

Are you sure that Borel-Cantelli requires a finite measure space? Wikipedia says it doesn't, and the proof there makes sense to me.

## @saz 2018-08-12 14:38:37

@MatthewKvalheim You are right; it should work without this assumption. I thought the finiteness would be necessary to use the continuity of the measure, but it isn't.