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. This follows because $L^{1}$ convergence means $f_{n}\to f$ in measure, and that in turn sugggests there exists a subsequence $f_{n_{k}}$ such that $f_{n_{k}}\to f$ pointwise a.e.
• $f_{n}\to f$ 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 $L^{1}$ convergence implies convergence 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 $\lim _{n\to \infty }\mu (E_{n,\epsilon })=0$