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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?

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  • $\begingroup$ 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. $\endgroup$ – Joshua Pepper Apr 10 '14 at 11:47
  • $\begingroup$ Well any bit of verification helps and I'm isolated from my math buddies at the moment. Thanks :) $\endgroup$ – Calvin Khor Apr 10 '14 at 12:08
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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 $f$.

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