Background Information:

Theorem 2.30 - Suppose that $\{f_n\}$ is Cauchy in measure. Then there is a measurable function $f$ such that $f_n\rightarrow f$ in measure, and there is a subsequence $\{f_{n_j}\}$ that converges to $f$ a.e. Moreover, if also $f_n\rightarrow g$ in measure, then $g=f$ a.e.


Exercise 33 - If $f_n\geq 0$ and $f_n\rightarrow f$ in measure then $\int f\leq \liminf \int f_n$

Attempted proof - Let $f_n\geq 0$ and $f_n\rightarrow f$ in measure then for every $\epsilon >0$ $$\mu\left(\{x:|f_n(x) - f(x)|\geq\epsilon\}\right)\rightarrow 0 \ \ \text{as} \ \ n\rightarrow \infty$$ Thus by Theorem 2.30 there is subsequence $\{f_{n_j}\}$ which converges to $f$ a.e. Then by Fatou's lemma $$\int f \leq \int \liminf f_{n_j} \leq \liminf \int f_{n_j}$$

I am not sure where to go from here any suggestions is greatly appreciated


@Wolfy, The idea in your attempt proof is correct but it fails basically because when we take a subsequence $\{f_{n_j}\}$ of $\{f_n\}$, we have
$$\liminf \int f_n \leq \liminf \int f_{n_j}$$

The "trick" is start by peeking a subsequence $\{f_{n_i}\}_i$ such that $$\int f_{n_i} \to \liminf_{n \to \infty} \int f_n $$ in other words: $$\lim_{i \to \infty}\int f_{n_i} = \liminf_{n \to \infty} \int f_n $$

Here is the proof in details.

Exercise 33 - If $f_n\geq 0$ and $f_n\rightarrow f$ in measure then $\int f\leq \liminf \int f_n$


Since $\{ \int f_n \}_n$ is just a sequence of complex number, there is subsequence $\{f_{n_i}\}_i$ such that $$\int f_{n_i} \to \liminf_{n \to \infty} \int f_n $$

Since $f_n\rightarrow f$ in measure, we have that $f_{n_i}\rightarrow f$ in measure. So there is a subsequence $\{f_{n_{i_j}}\}_j$ of $\{f_{n_i}\}_i$ such that $$f_{n_{i_j}} \to f \textrm{ a.e.} \tag{1} $$

Since $\{f_{n_{i_j}}\}_j$ is a subsequence of $\{f_{n_i}\}_i$, we also have $$\int f_{n_{i_j}} \to \liminf_{n \to \infty} \int f_n \tag{2}$$

So we have from $(1)$ and $(2)$, using Fatou's lemma,

$$ \int f = \int \lim_{j \to \infty } f_{n_{i_j}} = \int \liminf_{j \to \infty } f_{n_{i_j}}\leq \liminf_{j \to \infty } \int f_{n_{i_j}}= \lim_{j \to \infty } \int f_{n_{i_j}}= \liminf_{n \to \infty} \int f_n$$


Just do it in two steps. First pass to a subsequence $f_{n_j}$ so that $$\int f_{n_j} \to \liminf \int f_n$$ then pass to a further subsequence $f_{n_{j_\ell}}$ which converges to $f$ pointwise a.e. (possible since $f_{n_j}$ still converges to $f$ in measure) and then apply Fatou's Lemma.


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