[測度論] Dominated Convergence Theorem

Theorem: Dominated Convergence Theorem DCT  (real-valued functions)
令 sequence $\{f_n\} \subset L^1$ 滿足
(a) $f_n \to f$ almost everywhere
(b) 存在非負函數 $g \in L^1$ 使得 $|f_n| \leq g$ almost everywhere for all $n$
則 $$f \in L^1$$ 且
$$\lim_n \int f_n =\int f$$
Proof:  先證 $f \in L^1$：由於 $f_n \to f$ almost everywhere  且 $f_n \in L^1$，可知 $f$ measurable (see Proposition 2.11/2.12 )。由於 $|f_n| \leq g$ almost everywhere 故 $|f| \leq g$ almost everywhere，故 $f \in L^1$。

接著我們證 $\lim_n \int f_n =\int f$：注意到 $|f_n| \leq g$ almost everywhere，故我們有
$$-g \leq f_n \leq g \text{ almost everywhere}$$ 換言之，我們有
$$f_n + g \geq 0\text{ almost everywhere}$$ 與
$$g-f_n \geq 0 \text{almost everywhere}$$ 也就是說 $\{f_n + g\}, \{g-f_n\} \in L^+$。由 Fatou Lemma 我們有
$$\begin{gathered}   \int {\mathop {\lim \inf }\limits_n } ({f_n} + g) \leqslant \lim \inf \int {({f_n} + g)} ; \hfill \\   \int {\mathop {\lim \inf }\limits_n } (g - {f_n}) \leqslant \lim \inf \int {(g - {f_n})}  \hfill \\ \end{gathered} \;\;\; (*)$$ 注意到上述積分不等式左方，積分項有以下結果 (利用 limsup or liminf 的性質)，我們有 $\mathop {\lim \inf }\limits_n \left( {{f_n} + g} \right) = g + \mathop {\lim \inf }\limits_n {f_n}$ 以及 $$\mathop {\lim \inf }\limits_n \left( {g - {f_n}} \right) = g + \mathop {\lim \inf }\limits_n \left( { - {f_n}} \right) = g - \mathop {\lim \sup }\limits_n {f_n}$$ 同理，對於積分不等式右方，我們有
$$\lim \inf \int {({f_n} + g)}  = \lim \inf \left( {\int {{f_n} + \int g } } \right) = \lim \inf \int {{f_n} + \int g }$$ 與
$$\lim \inf \int {(g - {f_n})  = \int {g + \lim \inf \left( { - \int {{f_n}} } \right)}  = \int {g - \lim \sup \int {{f_n}} } }$$
將此結果帶入 $(*)$ 我們有
$$\begin{align*}   &\left\{ \begin{gathered}   \int {\mathop {\lim \inf }\limits_n } ({f_n} + g) \leqslant \lim \inf \int {({f_n} + g)} ; \hfill \\   \int {\mathop {\lim \inf }\limits_n } (g - {f_n}) \leqslant \lim \inf \int {(g - {f_n})}  \hfill \\ \end{gathered}  \right. \hfill \\    &\Rightarrow \left\{ \begin{gathered}   \int g  + \int {\mathop {\lim \inf }\limits_n } {f_n} \leqslant \lim \inf \int {{f_n} + \int g } ; \hfill \\   \int g  - \int {\mathop {\lim \sup }\limits_n {f_n}}  \leqslant \int {g - \lim \sup \int {{f_n}} }  \hfill \\ \end{gathered}  \right. \hfill \\    &\Rightarrow \left\{ \begin{gathered}   \int {\mathop {\lim \inf }\limits_n } {f_n} \leqslant \lim \inf \int {{f_n}} ; \hfill \\    - \int {\mathop {\lim \sup }\limits_n {f_n}}  \leqslant  - \lim \sup \int {{f_n}}  \hfill \\ \end{gathered}  \right. \hfill \\ \end{align*}$$ 注意到 $f = \lim_n f_n = \liminf_n f_n = \limsup_n f_n$ ，而且 $\{x: \lim_n f_n \neq f(x)\}$ 的測度為 $0$，故 $\int f = \int \liminf f_n  = \int \limsup f_n$。 現在我們進一步改寫上式
$$\begin{gathered}   \left\{ \begin{gathered}   \int f  \leqslant \lim \inf \int {{f_n}} ; \hfill \\   \int f  \geqslant \lim \sup \int {{f_n}}  \hfill \\ \end{gathered}  \right. \hfill \\    \Rightarrow \lim \sup \int {{f_n} \leqslant \int {f \leqslant \lim \inf \int {{f_n}} } }  \hfill \\ \end{gathered}$$ 亦即 $\int f_n \to \int f$。至此得證。$\square$



Remarks:
$\liminf (a_k + b+k) = \liminf a_k + \liminf b_k$ 不全為然。Counter example?

上述等式成立僅僅 $\lim a_k = a$ 則 $\liminf_k (a_k + b_k) = a+\liminf b_k$。