# If $(f_j)\to f$ in measure, show that $\int_X f\,d\mu\le\varliminf\int_X f_j\,d\mu$

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.

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