Set of Points at which Sequence of Measurable Functions does not Converge to Given Measurable Function is Measurable
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Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \R$ be a $\Sigma$-measurable function.
For each $n \in \N$, let $f_n : X \to \R$ be a $\Sigma$-measurable function.
Then:
- $\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x}$ is $\Sigma$-measurable.
Proof
From Expression for Set of Points at which Sequence of Functions does not Converge to Given Function, we have:
- $\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x} = \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$
From Pointwise Difference of Measurable Functions is Measurable, we have:
- $f_n - f$ is $\Sigma$-measurable for each $n \in \N$.
From Absolute Value of Measurable Function is Measurable, we have:
- $\size {f_n - f}$ is $\Sigma$-measurable for each $n \in \N$.
From Characterization of Measurable Functions, we have that:
- $\ds \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable for each $k, n \in \N$.
Since $\sigma$-Algebras are closed under countable union, we have:
- $\ds \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable for each $k, N \in \N$.
Then, from Sigma-Algebra Closed under Countable Intersection:
- $\ds \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable for each $k \in \N$.
Again, since Since $\sigma$-Algebras are closed under countable union, we have:
- $\ds \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ is $\Sigma$-measurable.
So:
- $\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x}$ is $\Sigma$-measurable.
$\blacksquare$