Prove Reverse Fatou's lemma

Question

Question

I am trying to prove the Reverse Fatou's lemma but I can't seem to get it. The statement is the following:

Suppose that $(f_n)_{n \in \mathbb{N}}$ is a sequence of measurable functions and $g$ an integrable function such that $f_n \leq g$ for all $n \in \mathbb{N}$. Then, $\limsup _{n \rightarrow \infty} \mu (f_n) = \mu ( \limsup _{n \rightarrow \infty} f_n)$ where $\mu$ is the integral on a specified measure space.

Attempt

We have a sequence $\lbrace f_k \rbrace$ in $\mathbb R$ and $E\subset \mathbb R$. We know that $\limsup f_k = \lim\limits_{j\rightarrow \infty} g_j$ where $g_j = \sup\limits_{k\geq j } f_k$. Thus we get that $$f_k \leq g_j \Rightarrow \int_E f_k \leq \int_E g_j \implies \sup\limits_{k\geq j }\int_E f_k \leq \int_E g_j $$ Taking the limit of both sides yields $$\lim\limits_{j\rightarrow \infty}\sup\limits_{k\geq j }\int_E f_k \leq \lim\limits_{j\rightarrow \infty}\int_E g_j $$ Which is equivalent to $$\limsup\limits_{j\rightarrow \infty}\int_E f_k \leq \liminf\limits_{j\rightarrow \infty}\int_E g_j $$

This is where I get stuck. I want to use Fatou's lemma but it won't work in this case. Is there a better way to prove this?

Answer 1

As $(f_n)_{n \in \mathbb{N}}$ and $g$ are both measurable, we know that $(g-f_n)$ is also measurable.

Therefore by Fatou's Lemma

$$\mu (\liminf _{n \rightarrow \infty} (g-f_n)) \leq \liminf _{n \rightarrow \infty}\mu(g-f_n) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space(1)$$

As the function $g$ is independent of $n$, we can rewrite $(1)$ as the following (by linearity of the integral)

$$ \mu (g) + \mu(\liminf _{n \rightarrow \infty} (-f_n)) \leq \mu(g)+ \liminf _{n \rightarrow \infty} (-\mu(f_n))\space \space \space \space \space \space \space \space \space (2)$$

We can immediately observe that $\mu (g)$ is present on both sides of the inequality which allows us to simplify to the following result:

$$ \mu(\liminf _{n \rightarrow \infty} (-f_n)) \leq \liminf _{n \rightarrow \infty} (-\mu(f_n))\space \space \space \space \space \space \space \space \space (3)$$

We now complete the proof by using the fact that $\liminf _{n\rightarrow \infty} (-f_n) = - \limsup _{n \rightarrow \infty} (f_n)$ and substituting this into $(3)$ to deduce

$$ -\mu(\limsup _{n \rightarrow \infty}(f_n)) \leq -\limsup _{n \rightarrow \infty} (\mu(f_n))\space \space \space \space \space \space \space \space \space (4)$$

And by rearranging inequality $(4)$ above, we get the final result:

$$ \fbox{$\limsup _{n \rightarrow \infty} (\mu(f_n)) \space \leq \space \mu(\limsup _{n \rightarrow \infty}(f_n))$} $$

Note: as discussed in the comments, equality does not hold here unless the limit exists and the hypotheses for monotone convergence are satisfied