# Monotone Convergence Theorem

1. Apr 29, 2014

### BrainHurts

1. The problem statement, all variables and given/known data

Let f be a non-negative measurable function. Prove that

$\lim _{n \rightarrow \infty} \int (f \wedge n) \rightarrow \int f.$

3. The attempt at a solution

I feel like I'm supposed to use the monotone convergence theorem.

I don't know if I'm on the right track but I created a sequence of functions so that

$h_1(x) \leq h_2(x) \cdots$ where

$h_1(x) = \min(f_1(x), n)$

$h_2(x) = \min(f_2(x),n)$

$\vdots$

$h_n(x) = \min(f_n(x),n)$

So the $h(x) = \lim_{n\rightarrow\infty} h_n(x) = \lim_{n\rightarrow \infty} \min(f_n,n) = \lim_{n \rightarrow \infty}\min(f,n)$

2. Apr 29, 2014

### Dick

I'm guessing $f \wedge n$ means min(f,n)? Then why not just define $f_n=\min(f,n)$? What do YOU mean by your $f_n$??

3. Apr 29, 2014

### BrainHurts

yes $f \wedge n = \min(f, n)$

so define

$f_1(x) = \min(f(x),1)$

$f_2(x) = \min(f(x),2)$

$\vdots$

$f_n(x) = \min(f(x),n)$

In short we still have that $f_1(x) \leq f_2(x) \leq \cdots \leq f_n(x)$ for all x.

Well I want $\lim_{n \rightarrow \infty}f_n \rightarrow f$ for all x because if I have this situation I can use the monotone convergence theorem.

I'm just not sure if that's the case.

Last edited: Apr 29, 2014
4. Apr 29, 2014

### Dick

I think it's all fine, except $f_1(x) = \min(f_1(x),1)$ doesn't do a good job of defining $f_1$. $f_1(x) = \min(f(x),1)$ is much better. Are you asking why $f_n \rightarrow f$?

Last edited: Apr 30, 2014
5. Apr 29, 2014

### BrainHurts

Sorry I just made some edits, I saw my mistake.

6. Apr 30, 2014

### BrainHurts

I think I'm covered, because I'll have to consider the case when f is finite and f is infinite correct?

7. Apr 30, 2014

### Dick

Maybe. f is nonnegative. So either the integral exists or it's '+infinity'. I think in either case the sequence convergence is correct. If you allow things like '+infinity'. Think about a function like f(x)=1/x^2. Define f(0) however you want.

Last edited: Apr 30, 2014