Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Monotone convergence theorem doesn't require the sequence of functions $f_n$'s to be $L^1$. When $f_n\in L^1$, will its pointwise limit function $f$ also be in $L^1$? Thanks!

share|improve this question

2 Answers 2

up vote 5 down vote accepted

Not necessarily.

Define $f_n$ as follows.

On $[-n,n]$, set $f_n(x):=1$ and $f_n(x):=0$ elsewhere.

Each $f_n$ is in $L^1$.

But the pointwise limit, which is equal to the constant function $f(x)=1$ on $\mathbb{R}$, is not in $L^1$.

share|improve this answer

Not quite, since the monotone convergence theorem includes the case where $\lim_{n \to \infty} \int f_n \ d\mu = \infty$. But if the limit is finite, $f \in L^1$.

share|improve this answer
    
Thank you! Why is $\lim_{n→∞}∫f_n dμ=∞$, if $f_n \in L^1$? –  Ethan Jan 27 '13 at 19:04
    
Why not? @julien provided one example, it's easy to find many others. –  Robert Israel Jan 27 '13 at 19:16

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.