# $-\liminf _n |f-f_n|d\mu=\limsup_n |f-f_n|d\mu$, is this actually a true claim?

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 ?

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 $$\int \liminf g_n\, d \mu = 2\int |f|\, d \mu \leq \liminf \int g_n\, d \mu = 2\int |f| d \mu+\liminf\left(- \int |f-f_n|\,d\mu\right) \\ =2\int |f| d \mu-\limsup \int |f-f_n|\,d\mu$$
• 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. Feb 6 '19 at 22:28