> [!definition] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]]/[[Probability|probability]] space and $\seq{f_n}$ be a sequence of [[Measurable Function|measurable functions]]/[[Random Variable|random variables]]. The sequence **converges in measure/probability** to $f$, denoted as $f_n \xrightarrow{\mu/\bp} f$, if for all $\varepsilon > 0$, > $ > \limv{n}\mu\bracs{\abs{f_n - f} > \varepsilon} = 0 > $ > [!theorem] > > Let $(X, \cm, \mu)$ be a finite measure space, and $M(X, \cm, \mu)$ be the space of real-valued measurable functions that are finite [[Almost Everywhere|almost everywhere]]. For each $f \in M(X, \cm, \mu)$, define > $ > \norm{f} = \int \frac{\abs{f}}{1 + \abs{f}} > $ > then $\norm{\cdot}$ is a [[Quasi-Norm|quasi-norm]] on $M(X, \cm, \mu)$, which induces the topology of convergence in measure. > > *Proof*. By the [[Dominated Convergence Theorem]], $\norm{\lambda f_n} \to 0$ and $\norm{\lambda_n f} \to 0$ whenever $f_n \to 0$ or $\lambda_n \to 0$. Let $\seq{f_n} \subset M(X, \cm, \mu)$ such that $f_n \to 0$ in measure, then for any $\eps > 0$, > $ > \begin{align*} > \norm{f_n} &\le \int_{\bracs{\abs{f_n} > \eps}}\frac{\abs{f}}{1 + \abs{f}} + \int_{\bracs{\abs{f_n} \le \eps}}\frac{\abs{f}}{1 + \abs{f}} \\ > &\le \mu(\bracs{\abs{f_n} > \eps}) + \eps\mu(X) \to \eps\mu(X) > \end{align*} > $ > as $n \to \infty$. As the above holds for all $\eps > 0$, convergence in measure implies convergence in the quasi-norm. On the other hand, by [[Markov's Inequality]], > $ > \mu\bracs{\frac{\abs{f_n}}{1 + \abs{f_n}} > \eps} \le \frac{\norm{f_n}}{\eps} \to 0 > $ > as $n \to \infty$. > [!theorem] > > Let $\seq{f_n} \subset L^1$. If $f_n \to f$ in $L^1$, then $f_n \to f$ in measure. > > *Proof*. Let $\varepsilon > 0$ and > $ > E_n = \bracs{x: \abs{f_n(x) - f(x)} \ge \varepsilon} > $ > then > $ > \int \abs{f_n - f} \ge \int_{E_n}\abs{f_n - f} \ge \varepsilon \mu(E_n) > $ > which tends to $0$ as $n \to \infty$. > [!theorem] > > Suppose that $\mu$ is finite. Let $\seq{f_n}$ be a sequence of measurable functions and suppose that $f_n \to f$ [[Pointwise Convergence|pointwise]] [[Almost Everywhere|a.e.]], then $f_n \to f$ in measure. > > *Proof*. Consider the set > $ > E_k = \bracs{x: \sup_{n \ge k}\abs{f_n(x) - f(x)} \ge \varepsilon} > $ > Since for any $x$ (except in a null set), $f_n(x) \to f(x)$ pointwise, there exists $k$ large enough such that $x \not\in E_k$. Therefore $\bigcap_{k \in \nat}E_k$ is a [[Null Set|null set]]. We can now establish the inequalities > $ > \begin{align*} > \mu\bracs{\abs{f_n - f} \ge \varepsilon} &\le \mu(E_n) \\ > \limv{n}\mu\bracs{\abs{f_n - f} \ge \varepsilon} &\le \limv{n}\mu(E_n) = 0 > \end{align*} > $ > Therefore $f_n \to f$ in measure. > [!theorem] > > Let $\seq{f_n}$ be a sequence of measurable maps such that $f_n \to f$ in measure, then there exists a subsequence $\seq{f_{n_k}}$ such that $f_{n_k} \to f$ [[Almost Everywhere|a.e.]] pointwise. > > *Proof*. For any $k \in \nat$, choose $n_k$ such that > $ > \mu\bracs{\abs{f_n - f} > \frac{1}{k}} \le \frac{1}{2^k} > $ > Then for any $m \in \nat$, > $ > \sum_{k \ge 1}\mu\bracs{\abs{f_{n_k} - f} > \frac{1}{m}} > $ > the above sum converges due to our setup. By the [[Borel-Cantelli Lemmas|First Borel-Cantelli Lemma]], > $ > \mu\paren{\limsup_{k \to \infty}\bracs{\abs{f_{n_k} - f} > \frac{1}{m}}} = 0 > $ > Therefore $f_{n_k} \to f$ pointwise a.e. > [!theorem] > > Let $\seq{f_n}$ be a sequence of measurable functions. The sequence is **Cauchy** in measure if > $ > \mu(\bracs{x: \abs{f_m(x) - f_n(x)} < \varepsilon}) \to 0 > $ > as $m, n \to \infty$. If $f_n$ is Cauchy in measure, then there exists $f$ such that $f_n \to f$ in measure, and there exists a subsequence $\seq{f_{n_k}}$ that converges to $f$ a.e. Moreover, the above limit is unique [[Almost Everywhere|a.e.]]. > > *Proof*. Let $\seq{f_{n_k}}$ such that > $ > \mu(\bracs{x: \abs{f_{n_k}(x) - f_{n_{k + 1}}(x)}\ge 2^{-k}}) \le 2^{-k} > $ > Denote $E_n = \bracs{x: \abs{f_{n_k}(x) - f_{n_{k + 1}}(x)}\ge 2^{-k}}$, then for any $j \in \nat$, $F_j = \bigcup_{n \ge j}E_n$ satisfies $\mu(F_j) \le 2^{1 - j}$. If $x \not\in F_k$, then > $ > \begin{align*} > \abs{f_{n_j}(x) - f_{n_i}(x)} &\le \sum_{l = j}^{i - 1}\abs{f_{n_{l + 1}}(x) - f_{n_{l}}(x)} \\ > &\le \sum_{l = j}^{i - 1}2^{-l} \le 2^{1 - j} > \end{align*} > $ > for any $i \ge j \ge k$. Therefore $f_{n_k}$ converges pointwise on $F_k^c$. Let $F = \bigcap_{k \in \nat}F_k = \limsup E_k$, then $\mu(F) = 0$. Therefore $f_{n_k}$ converges to a measurable function $f$ a.e. > > Since $\seq{f_n}$ is Cauchy in measure, $f_n \to f$ in measure as well. As the limits cannot differ by more than any $\varepsilon > 0$ on any set of measure greater than $0$, the limit is unique a.e.