Ex4-18

Question. If f is measurable on E, define \omega_f(a) = |\{f > a\}| = |\{\mathbf{x} \in E \mid f(\mathbf{x}) > a\}| for -\infty < a < \infty. 
(1) If f_k \nearrow f, show that \omega_{f_k} \nearrow \omega_f. 
(2) If f_k \overset{m}{\longrightarrow} f, show that \omega_{f_k} \to \omega_f at each point of continuity of \omega_f.
Answer.
(1) Since f_k \nearrow f, we have
\omega_{f_k} = |\{\mathbf{x} \in E \mid f_k(\mathbf{x}) > a\}| \le |\{\mathbf{x} \in E \mid f_{k+1}(\mathbf{x}) > a\}| = \omega_{f_{k+1}}.
We claim that
\bigcup_{k \in \Z^+} \{\mathbf{x} \in E \mid f_k(\mathbf{x}) > a\} = \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\}.
First it is clear that \displaystyle \bigcup_{k \in \Z^+} \{\mathbf{x} \in E \mid f_k(\mathbf{x}) > a\} \sube \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\} since f_k \nearrow f. On the other hand, let \mathbf{x} \in \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\}. Since f_k \nearrow f, for all \varepsilon > 0, there exists K \in \Z^+ such that for all k \ge K
|f(\mathbf{x}) - f_k(\mathbf{x})| < \varepsilon.
In particular, take \varepsilon = f(\mathbf{x}) - a. Then \varepsilon > 0 and 
f(\mathbf{x}) - f_K(\mathbf{x}) < f(\mathbf{x}) - a,
which implies that f_K(\mathbf{x}) > a. Thus, \mathbf{x} \in \{\mathbf{x} \in E \mid f_k(\mathbf{x}) > a\}. This shows that \displaystyle\{\mathbf{x} \in E \mid f(\mathbf{x}) > a\} \sube \bigcup_{k \in \Z^+} \{\mathbf{x} \in E \mid f_k(\mathbf{x}) > a\}, which then proves the claim.
Now we have
\{\mathbf{x} \in E \mid f_k(\mathbf{x}) > a\} \nearrow \bigcup_{k \in \Z^+} \{\mathbf{x} \in E \mid f_k(\mathbf{x}) > a\} = \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\}.
Therefore, by Theorem 3.26 (i), we conclude that \omega_{f_k}\nearrow \omega_f.
(2) Given \varepsilon > 0. Let
E_k = \{\mathbf{x} \in E \mid |f_k(\mathbf{x}) -f(\mathbf{x})| > \varepsilon\}.
Since f_k \overset{m}{\longrightarrow} f , for any \delta > 0, there exists K \in \Z^+ such that |E_k| < \delta for all k \ge K (by Exercise 16). Since 
f_k = (f_k - f) + f \implies |f_k| \le |f_k - f| + |f|,
we have
\{\mathbf{x} \in E \mid f_k(\mathbf{x}) > a\} \sube E_k \cup \{\mathbf{x} \in E \mid f(\mathbf{x}) > a - \varepsilon\}.
It follows by that
\omega_{f_k}(a) \le |E_k| + \omega_f(a - \varepsilon) < \delta + \omega_f(a - \varepsilon)
for all k \ge K. This means that \omega_f(a - \varepsilon) is an eventual upper bound of \{\omega_{f_k}(a)\}. Hence, 
\limsup_{k \to \infty} \omega_{f_k}(a) \le w_f(a - \varepsilon).
Analogously, since 
f = (f - f_k) + f_k \implies |f| \le |f - f_k| + |f_k|,
we have
\{\mathbf{x} \in E \mid f(\mathbf{x}) > a + \varepsilon\} \sube E_k \cup \{\mathbf{x} \in E \mid f_k(\mathbf{x}) > a\}.
It follows that
\omega_f(a + \varepsilon) \le |E_k|+ \omega_{f_k}(a) < \delta + \omega_{f_k}(a)
for all k \ge K. Thus,
\liminf_{k \to \infty} \omega_{f_k}(a) \ge \omega_f(a + \varepsilon).
Now, let a be a point of continuity of \omega_f. Taking limits as \varepsilon \to 0 in the inequalities above, we find that
\limsup_{k \to \infty} \omega_{f_k}(a) \le \omega_f(a) \le \liminf_{k \to \infty} \omega_{f_k}(a).
Since \displaystyle \liminf_{k \to \infty} \omega_{f_k}(a) \le \limsup_{k \to \infty} \omega_{f_k}(a) is always true, we conclude that
\lim_{k \to \infty} \omega_{f_k}(a) = \omega_f(a),
which proves the assertion.