# Modes of Convergence

## Relevant Definitions

Denote our measure space as $(X,{\mathcal {M}},\mu )$ . Note that a property p holds for almost every $x\in X$ if the set $\{x\in X:p{\text{ doesn't hold }}\}$ has measure zero.

• A sequence of measurable functions $\{f_{n}\}$ converges pointwise if $f_{n}(x)\to f(x)$ for all $x\in X$ .
• A sequence of measureable functions $\{f_{n}\}$ converges uniformly if $\sup _{x\in X}|f_{n}(x)-f(x)|\to 0$ .
• A sequence of measurable functions $\{f_{n}\}$ converges to $f$ pointwise almost everywhere if $f_{n}(x)\to f(x)$ for almost every $x$ , or $\mu (\{x:f(x)\neq \lim _{n\to \infty }f(x)\})=0$ .
• A sequence of measurable functions $\{f_{n}\}$ converges in $L^{1}$ if $\int |f_{n}-f|\to 0.$ check Convergence in Measure for convergence in measure.

## Relevant Properties 

• $f_{n}\to f$ through uniform convergence implies $f_{n}\to f$ through pointwise convergence, which in turn implies $f_{n}\to f$ pointwise a.e. convergence.
• $f_{n}\to f$ through $L^{1}$ convergence implies $f_{n}\to f$ through pointwise a.e convergence up to a subsequence.
• $f_{n}\to f$ Pointwise a.e. convergence, equipped with dominating function, implies $f_{n}\to f$ in $L^{1}$ . To see the proof and examples of why we need a dominating function, read Dominated Convergence Theorem.
• Convergence in Measure lists relationships between convergence in measure and other forms of convergence.

## Proof $f_{n}\to f,{\text{ in }}L^{1}$ implies $f_{n}\to f$ in measure

Fix $\epsilon >0,$ define $E_{n,\epsilon }=\{x:|f_{n}(x)-f(x)|\geq \epsilon \}$ We can see that that

$\int _{E_{n,\epsilon }}\epsilon \leq \int _{E_{n,\epsilon }}|f_{n}-f|\leq \int |f_{n}-f|.$ we can see by $L^{1}$ convergence that $\lim _{n\to \infty }\int |f_{n}-f|=0,$ meaning that $\lim _{n\to \infty }\int _{E_{n,\epsilon }}\epsilon =\lim _{n\to \infty }\epsilon \mu (E_{n,\epsilon })=0,$ or $\lim _{n\to \infty }\mu (E_{n,\epsilon })=\lim _{n\to \infty }\mu (\{x:|f_{n}(x)-f(x)|\geq \epsilon \})=0$ . Hence the proposition follows.

## Proof $f_{n}\to f$ through $L^{1}$ convergence implies $f_{n}\to f$ pointwise a.e convergence up to a subsequence.

$L^{1}$ convergence means $f_{n}\to f$ in measure. Because $f_{n}\to f$ in measure, there exists a subsequence $f_{n_{k}}$ such that $f_{n_{k}}\to f$ pointwise a.e.

## Example: $f_{n}\to f$ in measure does not imply convergence in $L^{1}$ Let $f_{n}:={\frac {1}{n}}1_{[0,n]}$ . Then $f_{n}\to 0$ in measure, but $f_{n}$ does not converge to 0 in $L^{1}$ . This is because $\int f_{n}d\lambda =1$ for all $n$ .