# Show that $\int \liminf f_n \leq \liminf \int f_n \leq \limsup \int f_n \leq \int \limsup f_n$

Let $g$ be a non-negative integrable function over $E$ and suppose $\{f_n\}$ is a sequence of measurable functions on $E$ such that for each $n$, $|f_n| \leq g$ a.e. on $E$. Show that $$\int \liminf f_n \leq \liminf \int f_n \leq \limsup \int f_n \leq \int \limsup f_n.$$

I know that this problem is an application of the Lebesgue dominated convergence theorem.

Any idea of how to go about it thanks, I am really having a hard time with this problem.

By possibly excising a set of measure 0 we can assume that $|f_n| \leq g$ holds on $E$.\ Let $$g_n =\inf\limits_{k\geq n} f_k \leq f_n \ then \ g_n \rightarrow \lim\limits_{n \to \infty} \inf f_n$$ Note that from $-g \leq |f_n| \leq g$ for all $n$ it also follows that $-g \leq g_n \leq g$ for all $n$ and thus $|g_n| \leq g$\ Using LDCT it follows that $$\int\limits_{E} \lim\limits_{n \to \infty} \inf f_n = \lim\limits_{n \to \infty} \int\limits_{E} g_n = \lim\limits_{n \to \infty} inf \int\limits_{E} g_n \leq \lim\limits_{n \to \infty} \int\limits_{E} f_n$$\ Also $$\int\limits_{E} \lim\limits_{n \to \infty} inf f_n \leq \lim\limits_{n \to \infty} \int\limits_{E} f_n \leq \int\limits_{E} \lim\limits_{n \to \infty} \sup f_n \ \ (*)$$ Similarly,Let $h_n =\sup\limits_{k \geq n} f_k \geq f_n$ then $h_n \rightarrow \lim\limits_{n \to \infty} \sup f_n$ and note that $|h_n| \leq g$\ Again, using LDCT we get $$\int\limits_{E} \lim\limits_{n \to \infty} \sup f_n = \lim\limits_{n \to \infty} \int\limits_{E} h_n = \lim\limits_{n \to \infty} \sup \int\limits_{E} g_n \geq \lim\limits_{n \to \infty} \int\limits_{E} f_n$$ Also $$\int\limits_{E} \lim\limits_{n \to \infty} \sup f_n \geq \lim\limits_{n \to \infty} \int\limits_{E} f_n \geq \int\limits_{E} \lim\limits_{n \to \infty} \inf f_n \ \ (**)$$ From (*) and (**) we have: $$\int\limits_{E} \lim\limits_{n \to \infty} inf f_n \leq \lim\limits_{n \to \infty} inf \int\limits_{E} f_n \leq \lim\limits_{n \to \infty} \sup \int\limits_{E} f_n \leq \int\limits_{E} \lim\limits_{n \to \infty} \sup f_n$$

Hint: Apply Fatou's Lemma to the sequences $g + f_n, g- f_n.$

• Would you be able to elaborate a bit further? Once Fatou's Lemma is applied to $g\pm f_n$, we get $\int_E\liminf_{n\to\infty} (g\pm f_n)\leq \liminf_{n\to\infty} \int_E (g\pm f_n)$ and $\limsup_{n\to\infty}\int_E (g+f_n)\leq \int_E\limsup_{n\to\infty} (g+f_n)$, but how to proceed from there? – Sujaan Kunalan Aug 2 '16 at 16:39
• You just need to verify that if $a_n$ is a sequence in $\mathbb R,$ then $\liminf (c+a_n) = c +\liminf a_n,$ where $c$ is a constant. – zhw. Aug 2 '16 at 17:18
• $\int \liminf (g\pm f_n)=\int( g\pm \liminf f_n)=\int g\pm\int \liminf f_n$ It is correct? – eraldcoil Dec 8 '18 at 4:30
• @eraldcoil Yes. – zhw. Dec 8 '18 at 16:27
• A doubt! $(f_n)$ is measurable but $g$ is only integrable. Why g is measurable? We need$(g\pm f_n)$ to be measurable in order to use Fatou's lemma ... Integrable implies measurable??? – eraldcoil Dec 11 '18 at 13:30

We can assume that $|f_n| \leq g$.

let $g_n=$inf $f_k < f_n$. $g_n \rightarrow$ lim inf $f_n$. Since $-g < f_n < g$ for all $n$, it follows that $-g < g_n < g$ thus $|g_n| \leq g$. Now by the lebesgue dominated convergence theorem we get $\int_{E}$ lim inf $f_n =$ lim $\int g_n =$ lim inf $\int_{E} g_n <$ lim inf $\int_{E} f_n$. by the monotonicity of the integral thus we have $\int_{E}$ lim inf $f_n \leq$ lim inf $\int_E f_n \leq$ lim sup $\int_E f_n$