# Measurable and Unif. Convergence in (a,b)

1. Jul 26, 2012

### Bacle2

Hi, All:

If {f_n}:ℝ→ℝ are measurable and f_n-->f pointwise, then convergence is a.e. uniform. Are there any conditions we can add to have f_n-->f in some open interval (a,b)?

Correction: convergence happens in some subset of finite measure; otherwise above not true.

Last edited: Jul 26, 2012
2. Jul 26, 2012

### mathman

Your original statement is incorrect. Example:

fn(x) = n2x for 0≤x≤1/n
...... = n - n2(x-1/n) for 1/n<x<2/n
...... = 0 otherwise.

fn(x) -> 0 pointwise, but is certainly not uniform, although it will be in any interval excluding [0,ε] for any ε.

3. Jul 26, 2012

### Bacle2

Yes, I was careless. I should have said: for any ε>0 , fn(x)→0 uniformly outside of a set of measure ε. Find Nε with 1/Nε<e. Then fn(x)→0 outside of [0,1/N).

Last edited: Jul 26, 2012
4. Jul 27, 2012

### mathman

The statement is confusing. What is the assumption and what are you trying to prove?

5. Jul 27, 2012

### Bacle2

The assumptions are the same: a sequence of measurable functions converge on a bounded subset of the real line. It follows, e.g., by Egorof's , that f_n-->f a.e. uniformly.

I am wondering if we can add some conditions , either on the functions, or on the domain/range of the functions of the sequence so that the functions converge uniformly _in some interval (a,b)_ , instead of just a.e., uniformly. Basically, we can remove a set of measure zero and get a 1st category space with empty interior.

6. Jul 27, 2012

### Bacle2

Never mind, thanks; I found the condition I needed.