# Convergence in measure of integrable functions implies limit is integrable?

I'm going through my handwritten notes for my upcoming exam (so not homework) and the above was stated but not proven in class. The full statement is a little different, but the above part is the only part I'm not sure about. Full statement below:

Suppose $f_n$ are $L^1(ℝ)$ and are Cauchy in measure. Then there is a subsequence $f_{n_k}$ convergent a.e. to a function $f∈ L^1$, and $f_n$ converges to $f$ in measure. Further if $f_n → g$ in measure, $f=g$ a.e.

I think I have a counterexample: $f_n(x) := \frac{1}{x}\mathbf{1}_{[-1,-1/n] ∪ [1/n,1]}(x)$. Each $f_n$ is $L^1$ and has integral $0$; they are Cauchy in measure because for $ε > 0$, \begin{align} \mu(|f_n - f_m| > ε ) \leq \mu(f_n ≠ f_m ) = \left|\frac{2}{n} - \frac{2}{m}\right| \xrightarrow[m,n→∞ ]{} 0\end{align} but the limiting function (a.e.) is $\frac{1}{x}\mathbf{1}_{[-1,1]}$ which is not integrable. Is this right?

• I'm fairly sure your example is correct (but I'm only just learning measure theory too, so take my opinion with a pinch of salt). However, if you assume that your sequence of functions is uniformly integrable, then the limit will be integrable. – Joshua Pepper Apr 10 '14 at 11:47
• Well any bit of verification helps and I'm isolated from my math buddies at the moment. Thanks :) – Calvin Khor Apr 10 '14 at 12:08

Indeed, the counter-example is correct. Like in this one, we reduce to a finite measure space. An almost everywhere convergent sequence $(f_n)_n$ has no reason to be bounded in $\mathbb L^1$. For example, we can take $f$ a non-integrable non-negative function and $(f_n)_n$ a sequence of simple functions increasing to $f$.