Convergence of measurable functions

The discussion of this article falls entirely into the category of standard Lebesgue measure on ${ℝ}^{𝑛}$, and all measurable functions and measurable spaces are Lebesgue measurable. For the difference and connection between Lebesgue measure theory and Borel measure theory, I refer you to Borel measure and Lebesgue measure.

2 Convergence a.e.

3 Almost uniform convergence

Notice that the definition of almost uniform convergence is different from uniform convergence a.e. because we require that it converges uniformly on non-measure-zero sets, despite that the set can be arbitrarily small. ¹

3.1 Egorov theorem

In mathematical analysis, the condition of uniform convergence is unusually strong, that in general you don’t have uniform convergence if you can only prove pointwise convergence.

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4 Convergence in measure

This kind of convergence coincide with that of probability, in probability we say that a sequence of random variables ${𝑋}_{𝑛}$ converges in probability to $𝑋$ if for all $𝜀>0$

$$\underset{𝑛\to \infty }{lim}\mathbb{P}\left(|{𝑋}_{𝑛}-𝑋|\ge 𝜀\right)=0$$

here the probability function $\mathbb{P}\left(𝑋\right)$ stands as a measure.

4.1 Properties of convergence in measure

We discuss briefly the convergence in measure compared to other kinds of convergence, first we have the following observation:

Let $𝐸=\left(0,\infty \right)$, define

$${𝑓}_{𝑛}\left(𝑥\right)=\{\begin{array}{c}0\text{if}𝑥\in \left(0,𝑛\right]\\ 1\text{if}𝑥\in \left(𝑛,\infty \right)\end{array}$$

we can verify that ${𝑓}_{𝑛}\stackrel{\text{a.e.}}{⟶}0$: Consider

$$\underset{𝑛\to \infty }{lim}𝜆\left(\left\{𝑥\in 𝐸|\left|{𝑓}_{𝑛}\left(𝑥\right)-0\right|\right\}\right)$$

Let $𝑥\in 𝐸$, we can find an $𝑁\in \mathbb{N}$ s.t. $𝑁>𝑥$, by definition of ${𝑓}_{𝑛}\left(𝑥\right)$, we have ${𝑓}_{𝑛}\left(𝑥\right)=0$ for all $𝑛\ge 𝑁$ (since now $𝑥\in \left(0,𝑛\right]$). Therefore ${𝑓}_{𝑛}$ converges pointwise to $0$.

But this function does not converges in standard Lebesgue measure, which requires

$$\forall 𝜀>0.\underset{𝑛\to \infty }{lim}𝜆\left(\left\{𝑥\in 𝐸|\left|{𝑓}_{𝑛}\left(𝑥\right)-0\right|\ge 𝜀\right\}\right)=0$$

However, notice that for each ${𝑓}_{𝑛}$, no matter how big $𝑛$ is, there are always uncountably many point in $\left(𝑛,+\infty \right)$ on which ${𝑓}_{𝑛}\left(𝑥\right)=1$, this shows that

$$\underset{𝑛\to \infty }{lim}𝜆\left(\left\{𝑥\in 𝐸|\left|{𝑓}_{𝑛}\left(𝑥\right)-0\right|\ge 1\right\}\right)=\infty$$

which shows that ${𝑓}_{𝑛}$ does not converges in measure to $𝑓$.

4.1.1 Lebesgue theorem

Let us pause and consider briefly why would this happen. An obvious reason is that $𝐸$ is infinitely big, so that the increment of $𝑛$ can never “fill” the rest of the space, so if we can somehow control the “rest” of the uncontrolled space as we advance, so that the unconstrolled space finally becomes zero, we may establish a stronger result, this is shown by Lebesgue theorem, Henri Lebesgue proved that under certain conditions, convergence a.e. implies convergence in measure.

4.1.2 Riesz theorem

With one direction we have another direction, Riesz theorem provides the method to turn convergence in measure into convergence a.e.

5 Relation diagram of three different kinds of convergence

Therefore, summarizing the above three different kinds, we can have the following diagram

1. ¹This is in particular a weaker version of convergence uniformly almost everywhere.
2. ²Notice that the inverse of Egorov theorem is automatically true (since uniform convergence is stronger), this makes Egorov theorem a necessary and sufficient condition for convergence a.u.