# $f$ converge in measure but fails to converge in $L^{1}$

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.

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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.