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This is the exercise 5 in page 109 of Analysis III of Amann and Escher.

Let $f_j,f:X\to\overline{\Bbb R}^+$ measurable functions such that $(f_j)\to f$ in measure. Show that

$$\int_Xf\,d\mu\le\varliminf\int_X f_j\,d\mu$$

Note: $X$ is a $\sigma$-finite space.

From a previous result I know that if $(f_j)\to f$ in measure then there is a subsequence $(f_{j_k})\to f$ almost everywhere.

My work so far: I set $g_j:=\inf_{k\ge j} f_j$. Then $(g_j)$ is increasing and by the existence of $(f_{j_k})$ we knows that $g_j\le f$ almost everywhere, thus WLOG we can consider that $g_j\le f$ in $X$.

Then by Fatou's lemma we have that $$ \int_X\varliminf f_j\,d\mu=\int_X \lim g_j\,d\mu\le\varliminf\int_X f_j\,d\mu\tag1 $$ Then if we shows that $(g_j)\to f$ almost everywhere we are done. However I was unable to prove it, I set $$ L:=\{x\in X: \lim |f(x)-g_j(x)|=0\}\\ A_{j,n}:=\{x\in X:|f(x)-f_j(x)|\ge 1/n\}\\ B_n:=\{x\in X:|f(x)-g_j(x)|\ge 1/n,\;\forall j\in\Bbb N\} $$ Then I tried to show that $L^\complement$ is a null set from the convergence in measure of $(f_j)$, trying to see if the $B_n$ are null, but I didnt find a way, and Im thinking that it is not necessarily true.

Then I started a different approach: if $\int_X f\le\varliminf\int_X f_j$ then eventually $\int_X f\le\int_X f_j$. This means that if $\int_X f=\infty$ then eventually $\int_X f_j=\infty$, and if $\int_X f=K<\infty$ then eventually $\int_X f_j\ge K$.

But from here I doesn't find something useful. Some help will be appreciated, thank you.

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There exists a subsequence $(f_{j_k})_{k \in \mathbb{N}}$ of $(f_j)_{j \in \mathbb{N}}$ such that

$$\liminf_{j \to \infty} \int_X f_j \, d\mu = \lim_{k \to \infty} \int_X f_{j_k} , d\mu. \tag{1}$$

Since $f_{j_k} \to f$ in measure, we can take a further subsequence $f_{j_{k_{m}}}$ which converges almost everywhere to $f$. Applying Fatou's lemma we get

$$\int_X f \, d\mu = \int_X \lim_{m \to \infty} f_{j_{k_m}} \, d\mu \leq \liminf_{m \to \infty} \int_X f_{j_{k_m}} \, d\mu \stackrel{(1)}{=} \liminf_{j \to \infty} \int_X f_j \, d\mu.$$

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