I was reading the solution of an exercise and saw this

$-\liminf _n |f-f_n|d\mu=\limsup_n |f-f_n|d\mu$

here $(E,\mathcal A,\mu)$ is a measure space, $(f_n)_{n \geq 1}$ is a sequence of measurable function that converges almost surely to $f$

and $\int_E |f_n|d \mu \to \int_E |f|d \mu $

the goal was to show that we have $\int_E |f_n-f|d \mu \to 0 $

for that they applied fatou's lemma to $g_n = |f_n|+|f| -|f-f_n|$

which gives

$$\int \liminf g_n d \mu = 2\int |f| d \mu \leq \liminf \int g_n d \mu = 2\int |f| d \mu-\liminf \int |f-f_n|d \mu$$

and at this point they concluded that $\limsup \int |f-f_n|d \mu = 0$ because $-\liminf _n |f-f_n|d\mu=\limsup_n |f-f_n|d\mu$

I honestly got lost at the last step, can someone shed some light on that ?


The following is true:

If $\{a_n\}_{n\in\mathbb N}\subset [-\infty,\infty]$, then $$ \liminf_{n\to\infty}\, (-a_n)=-\limsup_{n\to\infty} a_n. $$ Hence $$ \limsup_{n\to\infty} f_n=-\liminf_{n\to\infty}\, (-f_n). $$*

Hence, in the proof of the OP we have $$ \int \liminf g_n\, d \mu = 2\int |f|\, d \mu \leq \liminf \int g_n\, d \mu = 2\int |f| d \mu+\liminf\left(- \int |f-f_n|\,d\mu\right) \\ =2\int |f| d \mu-\limsup \int |f-f_n|\,d\mu $$

  • $\begingroup$ so $$0 \leq -\limsup |f-f_n| \leq 0$$ therefore $\limsup |f-f_n| = 0$, thanks I see now and it means there was a typo. $\endgroup$ Feb 6 '19 at 22:28
  • $\begingroup$ Correct - That's the idea $\endgroup$ Feb 6 '19 at 22:29

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