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The full question from this practice qual:

True/False: Let $(X,M,\mu)$ be any measure space. If $f_n,f \in L^1(\mu)$ are measurable functions, $f_n \rightarrow f$ $\mu$-a.e. and $\lim \int f_n \rightarrow \int f$, then $f_n \rightarrow f$ in $L^1(\mu)$.

I recognize that the question is essentially asking if I can move the limit inside the integral. The three theorems I have which allow me to do this are: Monotone Convergence Theorem, Fatou's Lemma, and Dominated Convergence Theorem. Since I don't have any increasing sequences of functions or dominating function, I would think if the answer is true, I need to use Fatou's Lemma. But I don't see any way to use Fatou's Lemma to justify it is true (the fact the question says "any" measure space signals to me it may be false, but I've been unable to construct a counter example)

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  • $\begingroup$ @PedroTamaroff So Fatou's Lemma says that $\int 2|f| \leq \liminf \int (|f|+|f_n|)-|f_n-f|$? $\endgroup$ – Brenton May 11 '16 at 3:05
  • $\begingroup$ It happens if and only if $\lim_{n \to \infty} \int_X |f_n| =\int_X |f|$. $\endgroup$ – Mambo May 11 '16 at 22:46
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False.

Take $f = 0$.

$f_n = \frac{1}{n}\chi_{[0, n]} -\frac {1}{n} \chi_{[-n, 0)}$.

$f_n \to f$ as $n \to \infty$ a.e.

But $\int |f_n| = 2$. So $f_n$ does not converge to $f$ in $L^1$.

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  • $\begingroup$ Doesn't this not satisfy the condition $\lim \int f_n \rightarrow \int f$? $\endgroup$ – Brenton May 11 '16 at 3:02
  • $\begingroup$ $\int f_n = 0$ right? $\endgroup$ – Shubhodip Mondal May 11 '16 at 3:02
  • $\begingroup$ (Note my comments shows this is true for positive functions, and this shows this is the only obstruction.) $\endgroup$ – Pedro Tamaroff May 11 '16 at 3:05
  • $\begingroup$ @ShubhodipMondal Ah I misread the $f_n$ function. So it's false in general but true if $f_n \geq 0$ a.e.? $\endgroup$ – Brenton May 11 '16 at 3:09
  • $\begingroup$ @brenton no it is not true even if the functions are non negative $\endgroup$ – clark May 11 '16 at 3:43
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For completeness sake: it is true if you assume that $|f_n|_1\to |f|_1$. Take $g_n = (|f|+|f_n|)-|f_n-f|$. By hypothesis $g_n\to 2|f|$ almost everywhere and $g_n\geqslant 0$, so Fatou says...?

This holds in general for $L^p$ spaces: if $f_n\to f$ almost everywhere and $|f_n|_p \to |f|_p$, then $|f-f_n|_p\to 0$. Again, taking $g_n= 2^{p-1}(|f_n|^p+|f|^p) - |f_n-f|^p$ gives by Fatou the result.

This is one of many results due to Riesz.

In particular, it is true if $f_n$ is positive almost everywhere.


Note the above gives a very easy proof the simple functions are dense in $L^p$ for any finite $p$: just manufacture a sequence of simple functions that tend pointwise and in norm to your function, which is easy!

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  • $\begingroup$ So Fatou's Lemma says that $\int 2|f| \leq \liminf \int (|f|+|f_n|)-|f_n-f|$? $\endgroup$ – Brenton May 11 '16 at 3:08
  • $\begingroup$ Ok and then I can isolate the $|f_n -f|$ term, and show it converges to $0$? $\endgroup$ – Brenton May 11 '16 at 3:09
  • $\begingroup$ You get that $\int 2 |f| \leqslant 2\int |f|-\limsup |f_n-f|_1 $; or that $\limsup |f_n-f|_1 \leqslant 0$! $\endgroup$ – Pedro Tamaroff May 11 '16 at 3:30
  • $\begingroup$ Haha, I made a mistake and was thinking $\lim \text{inf}\, |f_n - f| \le 0$. But I think even that is enough to say that $f_n \to f$ in $L^1$? $\endgroup$ – Shubhodip Mondal May 11 '16 at 3:35
  • $\begingroup$ @ShubhodipMondal That is the definition of $L^1$ convergence: if $\limsup x_n=0$ and $x_n$ is nonnegative, $x_n\to 0$. $\endgroup$ – Pedro Tamaroff May 11 '16 at 3:37

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