Convergence in Measure

Let $(X,{\mathcal {M}},\mu )$ denote a measure space and let $f_{n},f:X\to \mathbb {R}$ for $n\in \mathbb {N}$. The sequence $\{f_{n}\}_{n\in \mathbb {N} }$ converges to $f$ in measure if $\lim _{n\to \infty }\mu \left(\{x\in X:|f_{n}(x)-f(x)|\geq \epsilon \}\right)=0$ for any $\epsilon >0$. Furthermore, the sequence $\{f_{n}\}_{n\in \mathbb {N} }$ is Cauchy in measure if for every $\epsilon >0,$ $\mu (\{x\in X:|f_{n}(x)-f_{m}(x)|\geq \epsilon \})\to 0$ as $n,m\to \infty$

Properties

• If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_n \to f } in measure and $g_{n}\to g$ in measure, then $f_{n}+g_{n}\to f+g$ in measure.

• If $f_{n}\to f$ in measure and $g_{n}\to g$ in measure, then $f_{n}g_{n}\to fg$ in measure if $\mu (X)<\infty$.

• If $f_{n}\to f$ in measure and $g_{n}\to g$ in measure, then $f_{n}g_{n}\to fg$ in measure if this is a finite measure space.

Relation to other types of Convergence

• If $f_{n}\to f$ in $L^{1}(\mu )$ then $f_{n}\to f$ in measure

• If $f_{n}\to f$ in measure, then there exists a subsequence $\{f_{n_{k}}\}_{k\in \mathbb {N} }$ such that $f_{n_{k}}\to f$ almost everywhere.

• If $\mu (X)<\infty$ and $f_{n},f$ measurable s.t. $f_{n}\to f$ almost everywhere Then $f_{n}\to f$ in measure.