Let $f_n, f : X \longrightarrow \bar{\mathbb R} $ be measurable functions and $f^-$ integrable $(\bar {\mathbb R} := \mathbb R \cup \{+\infty, - \infty \})$. Show that $$f_n \geq f \text{ almost everywhere} \Rightarrow \int _X \lim_{n \rightarrow \infty} \inf f_n \text{ d}\mu \leq \lim_{n \rightarrow \infty} \inf \int _X f_n \text{ d}\mu $$

My attempt:

$\text{Case } 1$: $f \geq 0$

I can use Fatou's Lemma here

$\text{Case } 1$: $f \lt 0$

Since $f_n \geq f \Longrightarrow -f_n \leq f \Longrightarrow (f_n -f)_{_n} \text{ is a sequence of nonnegative functions}$.

Then $$\int _X \lim \inf f_n + \int -f \leq \int _X \lim \inf (f_n-f) \leq \lim \inf \int _X (f_n-f) \leq \lim \inf \int _X f_n + \int_X -f$$

$\Longrightarrow \int _X \lim \inf f_n \leq \lim \inf \int _X f_n $


Since I only know that $f^-$ is integrable can I follow that $|f|$ is integrable and therefore $f_n$ and $f$ are integrable since $f_n$ & $f$ are both measurable? Also I am not sure if my proof is correct so any help here is appreciated

  • 2
    $\begingroup$ Consider $f_n + f^-$ instead. $\endgroup$ – Daniel Fischer Nov 17 '17 at 20:26
  • $\begingroup$ Ohhh since $f \lt 0$ i can write $f = f^-$ right? $\endgroup$ – user391105 Nov 17 '17 at 20:28
  • 2
    $\begingroup$ No, $f^- = \max \{ - f, 0\} \geqslant 0$, so it would be $f = -f^-$ for $f < 0$. But $f_n + f^-$ works whether $f$ has constant sign or not, we always have $f_n + f^- \geqslant 0$ when $f_n \geqslant f$ (since $f_n + f^- \geqslant f + f^- = f^+ \geqslant 0$), and $f^-$ is assumed to be integrable, so we can cancel its integral from the inequality at the end. $\endgroup$ – Daniel Fischer Nov 17 '17 at 20:33
  • $\begingroup$ Yea right thank you @DanielFischer $\endgroup$ – user391105 Nov 17 '17 at 20:34
  • $\begingroup$ What is $f$? The limit function of $f_n$? Or just a lower bound? $\endgroup$ – Jack Nov 17 '17 at 20:35

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