# Set of Points at which Sequence of Measurable Functions does not Converge to Given Measurable Function is Measurable

## 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

$\ds \set {x \in X : \sequence {\map {f_n} x}_{n \in \N} \text { does not converge to } \map f x} = \bigcup_{k \mathop = 1}^\infty \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$
$f_n - f$ 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 $k, N \in \N$.
$\ds \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} \ge \frac 1 k}$ 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.

$\blacksquare$