# Modes of Convergence

## Relevant Definitions[1]

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

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

check Convergence in Measure for convergence in measure.

## Relevant Properties [2]

• ${\displaystyle f_{n}\to f}$ through uniform convergence implies ${\displaystyle f_{n}\to f}$ through pointwise convergence, which in turn implies ${\displaystyle f_{n}\to f}$ pointwise a.e. convergence.
• ${\displaystyle f_{n}\to f}$ through ${\displaystyle L^{1}}$ convergence implies ${\displaystyle f_{n}\to f}$ through pointwise a.e convergence up to a subsequence.
• ${\displaystyle f_{n}\to f}$ Pointwise a.e. convergence, equipped with dominating function, implies ${\displaystyle f_{n}\to f}$ in ${\displaystyle L^{1}}$. 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 ${\displaystyle f_{n}\to f,{\text{ in }}L^{1}}$ implies ${\displaystyle f_{n}\to f}$ in measure

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

${\displaystyle \int _{E_{n,\epsilon }}\epsilon \leq \int _{E_{n,\epsilon }}|f_{n}-f|\leq \int |f_{n}-f|.}$

we can see by ${\displaystyle L^{1}}$ convergence that ${\displaystyle \lim _{n\to \infty }\int |f_{n}-f|=0,}$ meaning that ${\displaystyle \lim _{n\to \infty }\int _{E_{n,\epsilon }}\epsilon =\lim _{n\to \infty }\epsilon \mu (E_{n,\epsilon })=0,}$

or ${\displaystyle \lim _{n\to \infty }\mu (E_{n,\epsilon })=\lim _{n\to \infty }\mu (\{x:|f_{n}(x)-f(x)|\geq \epsilon \})=0}$. Hence the proposition follows.

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

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

## Example: ${\displaystyle f_{n}\to f}$ in measure does not imply convergence in ${\displaystyle L^{1}}$

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

## References

1. Craig, Katy. MATH 201A Lecture 17. UC Santa Barbara, Fall 2020.
2. Craig, Katy. MATH 201A Lecture 18. UC Santa Barbara, Fall 2020.