---- > [!definition] Definition. ([[convergence in measure]]) > Let $(X, \Sigma ,\mu)$ be a [[measure|measure space]] and $f,f_{n} \ (n \in \mathbb{N}):X \to Y$ be [[measurable function|measurable functions]] into (say) a [[metric space]] $Y$. The [[sequence]] $(f_{n})$ is said to **converge globally in measure** if for all $\varepsilon>0$,[^1] $\lim_{n \to \infty} \mu \{ d(f, f_{n}) \geq \varepsilon \}=0.$ > The [[sequence]] $(f_{n})$ is said to **converge locally in measure** if for every $\varepsilon>0$ and every $E \in \Sigma$ with $\mu(E)<\infty$, $\lim_{n \to \infty} \mu \{ d\big( f |_{E}, f_{n} |_{E} \big) \geq \varepsilon \} =0.$ > On a [[finite measure|finite measure space]] the two notions are manifestly equivalent. Otherwise the precise meaning of the phrase "convergence in measure" must be determined from context. ^definition > [!specialization] > When $(X, \Sigma ,\mu)=(\Omega, \mathcal{F}, \mathbb{P})$ is a [[probability|probability space]], the term **convergence in probability** is used instead. ^specialization > [!note] Remark. > When $\mu(X) < \infty$, one has as a consequence of [[Egorov's Theorem]][^3] $\text{convergence a.e.} \implies \text{convergence in measure} .$ When $\mu(X)=\infty$ this no longer holds.[^2] The converse can fail even when $\mu(X)<\infty$,[^4] although one does have, for arbitrary $X$, that[^5]$\text{convergence in measure} \implies \text{some subsequence converges a.e.}$ > > > Also[^6], $L^{p} \text{ convergence }(\text{any }1\leq p \leq \infty) \implies \text{convergence in measure}.$ > (Note that, in turn, $L^{p}$ convergence implies a.e. convergence of some subsequence.) [^6]: Here $Y=E$ for $E$ a [[Borel set|Borel subset]] of $\mathbb{R}$. Suppose $f_{j} \to f$ in $L^{p}(E)$ for some $1 \leq p \leq \infty$, that is, $\lim_{j \to \infty} \|f-f_{j}\|_{p}=0$. Fix $\varepsilon>0$. [[Chebyshev's Inequality]] applied to $f-f_{j}$ with $t=\varepsilon$ gives $\mu \{ |f - f_{j}| \geq \varepsilon \} \leq \|f-f_{j}\|_{p}^{p} / \varepsilon^{p}.$ The RHS goes to zero as $j \to \infty$, thus so to does the LHS. [^5]: Suppose $(f_{n}) \to f$ in [[measure]]. Let $(\varepsilon_{k})$ be a [[sequence]] of positive numbers with $(\varepsilon_{k}) \downarrow 0$, e.g. $\varepsilon_{k}:= 1 / k$. For $k \in \mathbb{N}$, choose $n_{k}$ large enough that $\mu \{ d(f,f_{n_{k}}) \geq \varepsilon_{k}\}<2^{-k}.$ [[limit inferior and limit superior|Then]] $\begin{align} \mu( \bigcap_{k=1}^{\infty} \bigcup_{m \geq k} \{ d(f_{n_{m}}, f) \geq \varepsilon_{m} \})&= \lim_{k \downarrow \infty} \mu(\bigcup_{m \geq k} \{ d(f_{n_{m}}, f)\geq \varepsilon_{m} \}) \\ & \leq \lim_{k \to \infty} \left( \sum_{m \geq k} \underbrace{\mu \{ d(f_{n_{m}}, f)\geq \varepsilon_{m} \} }_{ \leq 2^{-m} } \right) \\ &=0,\end{align}$where the first equality used [[measure|continuity from above]] and the second used [[measure|countable subadditivity of measure]]. For the last equality, note that $\sum_{m \geq k} 2 ^{-m}=\overbrace{ \sum_{k=1}^{\infty} 2 ^{-k} }^{ =1 } - \overbrace{ \sum_{k=1}^{m-1} 2^{-k} }^{ = 1+2^{1-m} }=2 \cdot 2^{-m} \xrightarrow[m \to \infty]{}0$using the formulas for [[geometric series]] $\left( r=\frac{1}{2} \right)$ and [[finite geometric series]]. We therefore see that the set $\{ x \in X : x \in \{ d(f_{n_{k}}, f) \geq \varepsilon_{k} \} \text{ for infinitely many }k \}$has [[measure zero]]; thus for almost every $x \in X$ we have $d(f_{n_{k}}(x), f(x)) < \varepsilon_{k}.$Taking $k\to \infty$ sends the RHS to $0$ by definition of $(\varepsilon_{k})$, and so $\lim_{k \to \infty} d(f_{n_{k}}(x), f(x))=0$ for almost all $x$. The result follows. > [!proof] > ^proof [^2]: E.g. $f_{n}=1_{[n, n+1]}$ on $X=\mathbb{R}$ with the [[Lebesgue measure]] converges a.e. to $0$, but $\mu \left\{ x \in X : |f_{n}(x)-x| \geq \frac{1}{59} \right\} \equiv 1$. [^3]: Proof: suppose $(f_{n})$ converges a.e. to $f$. Fix $\varepsilon>0$. Let $\varepsilon'>0$. We want to find $N=N(\varepsilon')$ such that for all $n \geq N$, $\mu \{ d(f,f_{n}) \geq \varepsilon\}<\varepsilon'$. By [[Egorov's Theorem]] (which requires $\mu(X)<\infty$) there exists a [[σ-algebra|measurable subset]] $B \subset X$ such that $\mu(X-B)<\varepsilon'$ and $f_{n} \to f$ [[uniform convergence|uniformly]] on $B$. Obtain $N$ witnessing this for $\varepsilon$: obtain $N$ so that for all $n \geq N$ one has $d(f_{n}(x),f(x))<\varepsilon \text{ for all } x \in B.$Then for all $n \geq N$, $\mu( \{ x \in X : d(f_{n}(x), f(x)) \geq \varepsilon\})\leq \mu(X-B) < \varepsilon',$as desired. [^4]: Consider the [[typewriter sequence]] $(f_{n})$ on the [[interval]] $[0,1]$ endowed with the [[Lebesgue measure]]. $(f_{n})$ converges in [[measure]] to $0$, since the $f_{n}$ are [[characteristic function|characteristic function]] on successively smaller intervals (in particular, whose length goes to $0$ with $2^{-j}$). However, $(f_{n}(x))$ does not [[sequence|converge]] for any $x$, as the sequence hits each of $1$ and $0$ infinitely many times. [^1]: The notation $\{ d(f, f_{n}) \geq \varepsilon \}$ stands for the set $\{ x \in X: d(f(x), f_{n}(x)) \geq \varepsilon \}=\big(d \circ (f \times f_{n})\big)^{-1}\big([\varepsilon, \infty)\big)$, like in [[random variable|probability]]. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```