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.

I am thinking I can use the function $$f_{n}= n1_{[0,1/n]}$$ This will work beacuse $f_{n}\rightarrow 0 $ in measure because $$1/n \rightarrow 0$$ as $n$ gets bigger, but when we take the integral of $f_{n}$ over $[0,1]$ then we will clearly get 1 for all n, no matter larger or smaller, which will essentially mean the limit is not 0.

I want to see if my understanding is correct here?

What I always had in my mind is since the integral of $f_{n}$ is finite over $[0,1]$, I would simply think the integral is convergent. But in the problem above I simply do not have the limit 0. But the function still converge in $L^{1}$. What is wrong with my understanding? Or my example is completely out of track?

Can somebody pull me out of this confusion. Thank you in advance.

share|improve this question
add comment

1 Answer 1

up vote 5 down vote accepted

For a sequence of functions $f_n$ to converge in $L^1(\mathbb R)$, there must be some function $g$ such that $$\lim\limits_{n\to\infty} \int_{-\infty}^\infty |f_n(x)-g(x)|dx=0.$$ With $f_n=n1_{[0,1/n]}$ we have $$\begin{align} \int_{-\infty}^\infty |f_n(x)-g(x)|dx &= \int_{0}^{1/n} |n-g(x)|dx + \int_{\mathbb R\setminus [0,1/n]}|g(x)|dx\\\\ &\ge \int_{0}^{1/n} (n-|g(x)|)dx + \int_{\mathbb R\setminus [0,1/n]}|g(x)|dx\\\\ &\ge 1 -\int_0^{1/n} |g(x)|dx\\\\ \end{align}$$ and since the limit as $n\to \infty$ of this must be at most $0$, we have $$\liminf\limits_{n\to\infty} \int_0^{1/n}|g(x)|dx\ge 1$$ which is clearly impossible for any $L^1$ function.

share|improve this answer
add comment

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.