If $\{f_n\}$ is Cauchy in measure, then there is a measurable function $f$, such that $\{f_n\}$ converges in measure to $f$

Question

The theorem is from Real Analysis (Carothers). Let $\{f_n\}$ be a sequence of real valued measurable functions, all defined on a common measurable domain $D$. If $\{f_n\}$ is Cauchy in measure, then there is a measurable function $f:D\rightarrow \mathbb{R}$, such that $\{f_n\}$ converges in measure to $f$. Moreover, there is a subsequence $\{f_{n_{k}}\}$ that converges pointwise a.e. to $f$. And the proof is shown in the picture

Can someone explain the last line to me? The stuff in the red box. I don't quite understand how they get that inequality.

Answer 1

The previous inequality shows that

$$|f(x)-f_{n_k}(x)| < 2^{-k+1} \qquad \text{for all $x \in D \backslash \bigcup_{j=k}^{\infty} E_j$,}$$

i.e.

$$x \in D \backslash \bigcup_{j=k}^{\infty} E_j \implies x \in \{|f-f_{n_k}| < 2^{-k+1}\}.$$

This is equivalent to

$$D \backslash \bigcup_{j=k}^{\infty} E_j \subseteq \{|f-f_{n_k}| < 2^{-k+1}\}.$$

Taking complement on both sides yields

$$\{|f-f_{n_k}| \geq 2^{-k+1}\} \subseteq \bigcup_{j=k}^{\infty} E_j,$$

and now the monotonicity of the measures proves the claimed inequality.