📂Measure Theory

Yegorov's Theorem

Theorem^{1 2}

Let $(X, \mathcal{E}, \mu)$ be a measure space, called $\mu (X) \lt \infty$. Let $f$ be a measurable function, and $\left\{ f_{n} \right\}$ be a sequence of measurable functions that converge to $f$ almost everywhere.

$$f_{n} \to f \text{ a.e. }$$

Then, for every $\epsilon \gt 0$, there exists a $E \subset X$ that satisfies the following

$$\mu (E) \lt \epsilon \quad \text{ and } \quad f_{n} \rightrightarrows f \text{ on } E^{c}$$

Explanation

This theorem, in a nutshell, tells us that for measurable functions, pointwise convergence and uniform convergence are almost the same.

The convergence described in the theorem is also called almost uniform convergence. If the following holds, it is said that $f_{n}$ converges to $f$ almost uniformly.

$$\forall \epsilon \gt 0,\quad \exists E \subset \mathcal{E} \text{ such that } \mu (E) \lt \epsilon \text{ and } f_{n} \rightrightarrows f \text{ on } E^{c}$$

The following facts hold. If it converges almost uniformly,

• it converges almost everywhere.
• it converges in measure.

Proof

Without loss of generality, let’s assume $f_{n} \to f$ for all $x \in X$. For $k, n \in \N$, let’s set $E_{n}(k)$ as follows.

$$E_{n}(k) = \bigcup \limits_{m = n}^{\infty} \left\{ x : \left| f_{m}(x) - f(x) \right| \ge \frac{1}{k} \right\} \tag{1}$$

Then, for a fixed $k$ satisfying $E_{n+1}(k) \subset E_{n}(k)$ and since $f_{n} \to f$, the following holds.

$$\bigcap\limits_{n=1}^{\infty} E_{n}(k) = \varnothing$$

Since it is assumed that $\mu (X) \lt \infty$, $\mu (E_{n}(k)) \to 0 \text{ as } n \to \infty$ follows. Therefore, for the given $\epsilon \gt 0$ and $k \in \N$, select a sufficiently large $n_{k}$ such that $\mu \left( E_{n_{k}}(k) \right) \lt \epsilon 2^{-k}$. Let’s set it as $E = \cup_{k=1}^{\infty} E_{n_{k}}(k)$. Then $\mu (E) \lt \epsilon$.

Also, if $x \notin E$, for all $k$, $x \notin E_{n_{k}}(k)$, and by definition $(1)$, the following holds.

$$\left| f_{n}(x) - f(x) \right| \lt \frac{1}{k} \text{ for } n \gt n_{k}$$

Therefore, $f_{n} \rightrightarrows f \text{ on } E^{c}$.

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