21.1 Measurable Functions

Definition 21.1.1 (Measurable Function).label Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces and $f: X \to Y$ be a mapping, then $f$ is $(\cm, \cn)$-measurable if $f^{-1}(E) \in \cm$ for all $E \in \cn$.

Definition 21.1.2 (Borel Measurable).label Let $(X, \cm)$ be a measurable space, $Y$ be a topological space, and $f: X \to Y$ be a mapping, then $f$ is Borel measurable if $f$ is $(\cm, \cb_{Y})$-measurable.

Definition 21.1.3 (Convergence Almost Everywhere).label Let $(X, \cm, \mu)$ be a measure space, $Y$ be a topological space, $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_{n} \to f$ almost everywhere/a.e. if there exists a $\mu$-null set $N \in \cm$ such that $f_{n} \to f$ pointwise on $X \setminus N$.

Lemma 21.1.4.label Let $X, Y$ be topological spaces and $f: X \to Y$ be continuous, then $f$ is Borel measurable.

Lemma 21.1.5.label Let $(X, \cm)$ and $(Y, \cn = \sigma(\ce))$ be measurable spaces, and $f: X \to Y$ such that $f^{-1}(\ce) \subset \cm$, then $f$ is $(\cm, \cn)$-measurable.

Proof. Let $\mathcal{F}= \bracs{E \in \ce| f^{-1}(E) \in \cm}$.$\square$

Definition 21.1.6 (Generated $\sigma$-Algebra).label Let $X$ be a set, $\bracs{(Y_i, \cn_i)}_{i \in I}$ be measurable spaces, and $\seqi{f}$ with $f_{i}: X \to Y_{i}$ for each $i \in I$. The $\sigma$-algebra generated by $\seqi{f}$,

\[\sigma(\bracs{f_i| i \in I}) = \sigma\paren{f_i^{-1}(\cn_i)|i \in I}\]

is the smallest $\sigma$-algebra on $X$ such that each $\seqi{f}$ is measurable.

Theorem 21.1.7 (Egoroff).label Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_{n} \to f$ almost everywhere, then for any $\eps > 0$, there exists $E \in \cm$ such that:

  1. (1)

    $f_{n} \to f$ uniformly on $E$.

  2. (2)

    $\mu(X \setminus E) < \eps$

Proof. Since $f_{n} \to f$ almost everywhere, for any $\eps > 0$,

\[\mu\paren{\limsup_{n \to \infty}\bracs{d(f_n, f) > \eps}}= 0\]

By continuity from above (Proposition 18.1.5),

\[\lim_{N \to \infty}\paren{\bigcup_{n \ge N}\bracs{d(f_n, f) > \eps}}= 0\]

For each $k \in \natp$, let $N_{k} \in \natp$ such that

\[\paren{\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}}< 2^{-k}\eps\]

then by subadditivity,

\[\mu\paren{\bigcup_{k \in \natp}\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}}\le \eps\]

and $f_{n} \to f$ uniformly on

\[\braks{\bigcup_{k \in \natp}\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}}^{c}\]

$\square$