# convergence in measure implies convergence in mean?

Given a sequence of functions $\{f_1,f_2,\dots\}$ converging in measure to some measurable function $f$. Is it necessary true that they also converge in mean to $f$?

Convergence in mean is defined as follows $\int \lvert f_n-f \rvert$ goes to zero as $n$ goes to infinity.

I know that the converse of the above is true. Proof is as follows:

Given $\epsilon>0$, we know need to show that for each $\delta$, there exist $n_0$ such that $\mu(\{x\mid \lvert f(x)-f_n(x)\rvert\geq \epsilon\})<\delta$ for all $n\geq n_0$, where $\mu$ is the measure. Notice that $\mu(\{x\mid \lvert f(x)-f_n(x)\rvert\geq \epsilon\})\cdot \epsilon\leq \int_E \lvert f_n(x)-f(x)\rvert$. Since the sequence converge in mean, the right hand side goes to zero as $n$ goes to infinity, thus so does the measure.

Now, back to the original question. Intuitively, knowing my calculus 3 stuff, I could have a sequence of function converge to some measurable real-valued function, but that function is not integrable since it blows up maybe when $x=0$. But how do I give a counter example? Or do I have to use a direct proof?

Consider $\mathbb{R}$ with Lebesgue measure, and define $f_n(x)=\frac{1}{n}$ if $0\leq x\leq n$, $f_n(x)=0$ otherwise.
Then $f_n\to 0$ in measure, but $\int f_n=1$ for all $n$, so $f_n\not\to0$ in $L^1$.