Convergence of Measures

Definition ¹

Let’s assume that a measure space $( X , \mathcal{E} , \mu)$ is given.

A sequence of measurable functions $\left\{ f_{n} \right\}_{n \in \mathbb{N}}$ is said to converge in measure to some measurable function $f$ if it satisfies the following for all $M >0$. $$ \lim_{n \to \infty} \mu \left( \left\{ x \in X : | f_{n}(x) - f(x) | \ge M \right\} \right) = 0 $$ The sequence $\left\{ f_{n} \right\}_{n \in \mathbb{N}}$ is called Cauchy in Measure if it satisfies the following for all $M >0$. $$ \lim_{n,m \to \infty} \mu \left( \left\{ x \in X : | f_{m}(x) - f_{n}(x) | \ge M \right\} \right) = 0 $$

Description

In terms of probability theory, this is known as convergence in probability.

The definition of convergence neatly describes the convergence we think of. However, the reason for defining a new convergence by involving measure, despite its complexity, is that convergence can be excessively difficult. Nonetheless, if we can compromise to a degree where $f_{n}$ becomes sufficiently similar to $f$, such that the area not becoming similar measures to $\mu$ converges to $0$, then we can discuss more.

This is similar to almost everywhere in measure theory. Moreover, convergence in measure is a step back from almost everywhere, as seen by the following properties, showing how it can be used under very unfavorable conditions.

Basic Properties

[1]: If $f_{n}$ uniformly converges to $f$, it pointwise converges.
[2]: If $f_{n}$ pointwise converges to $f$, it converges almost everywhere.
[3]: If $f_{n}$ converges almost everywhere to $f$, it converges in measure.
[4]: If $f_{n}$ converges in $\mathcal{L}_{p}$ convergence to $f$, it converges in measure.

Summing up [1]~[3]:

Uniform convergence $\implies$ Pointwise convergence $\implies$ Almost everywhere convergence $\implies$ Convergence in measure

Proofs

[1]

By the definition of uniform convergence, for all $x \in X$ and all $\varepsilon > 0$, there exists $N \in \mathbb{N}$ that satisfies $n \ge N \implies |f_{n}(x) - f(x)| < \varepsilon$, therefore, $f_{n}$ pointwise converges to $f$.

■

[2]

$f_{n}$ pointwise converging to $f$ means for all points $x$ in $X$ except for $E = \emptyset$, each function’s value $f_{n} (x)$ converges to $f(x)$. Since this $\mu ( \emptyset ) = 0$, $f_{n}$ almost everywhere converges to $f$.

■

[3]

$f_{n}$ almost everywhere converging to $f$ means that for all points $x$ in $X$ except for some $E \in \mathcal{E}$ that satisfies $\mu ( E) = 0$, each function’s value $f_{n} (x)$ converges to $f(x)$. Regardless of how $M > 0$ is given, $$ \left\{ x \in X : | f_{n}(x) - f(x) | \ge M \right\} \subset E $$ and always, by the measure’s monotonicity, $$ \mu \left( \left\{ x \in X : | f_{n}(x) - f(x) | \ge M \right\} \right) \le \mu ( E ) = 0 $$ thus, $f_{n}$ converges in measure to $f$.

■

[4]

For $M > 0$, $$ \begin{align*} \int_{X} | f_{n} - f |^{p} d \mu &\ge \int_{\left\{ x \in X : | f_{n}(x) - f(x) | \ge M \right\}} |f_{n} - f |^{p} d \mu \\ &\ge \int_{\left\{ x \in X : | f_{n}(x) - f(x) | \ge M \right\}} M^{p} d \mu \\ &\ge M^{p} \mu \left( \left\{ x \in X : | f_{n}(x) - f(x) | \ge M \right\} \right) \end{align*} $$ since $f_{n}$ converges in $\mathcal{L}_{p}$ convergence to $f$, $\displaystyle \lim_{n \to \infty } \int_{X} | f_{n} - f |^{p} d \mu = 0$ and $M>0$, hence $$ \lim_{n \to \infty} \mu \left( \left\{ x \in X : | f_{n}(x) - f(x) | \ge M \right\} \right) = 0 $$ Thus, $f_{n}$ converges in measure to $f$.

■