# $f_n(x)\rightarrow f(x)$ almost everywhere

$$\lim_{n\rightarrow\infty} f_n(x)=f(x)\space\space a.e.\text{for all x in E}$$

My Question:

1. How do you interpret this? As you get to the tail of the function sequence, each tail sequence and the function disagree only on negligible sets?

2. Is there a name for this type of convergence other than what we call almost everywhere convergence?

3. Is there any relationship between the a.e. convergence and pointwise convergence or uniform convergence?

These are discussed in the context of constructing the Lebesgue integration theory.

Reference: $$\textit{Real Analysis: Measure Theory, Integration, and Hilbert Spaces}$$. Elias M. Stein, Rami Shakarchi. Princeton University Press, 2009.

The definition says that the set $$\{x\in E: f_n(x)$$ does not converge to $$f(x)\}$$ has measure zero. Obviously this is something a bit weaker than pointwise convergence. (pointwise convergence is when the set I defined is empty).

As for names: in probability theory this type of convergence is also called "almost sure convergence". The intuition there is that if you pick a point then with probability $$1$$ it is a point where the sequence converges to the limit random variable.

• Isn't the standard definition that there is a set $X$ such that $(f_n)_{n \in \mathbb{N}}$ converges pointwise to $f$ for all $x \in X$ and the measure of the complement of $X$ is $0$? This would only be equivalent to your statement in the case of a complete measure space (subsets of measure zero sets are again measurable). Commented Sep 24, 2019 at 19:35
• Your definition is strictly stronger than mine. In your's the set on which $(f_n)_{n \in \mathbb{N}}$ does not converge needs to be measurable and have measure zero, in mine it only needs to be contained in a measurable set which has measure zero. Commented Sep 24, 2019 at 19:44
• Oh, I get your point. But the set of points where we don't have convergence is measurable anyway, right? The set of points where we have convergence is $\cap_{0<\epsilon\in\mathbb{Q}}\cup_{n_0=1}^\infty\cap_{n=n_0}^\infty \{|f_n(x)-f(x)|<\epsilon\}$. This is a measurable set, hence the complement is measurable as well.
– Mark
Commented Sep 24, 2019 at 19:48
• Sure, but only if you assume the $(f_n)_{n \in \mathbb{N}}$ to be measurable. Commented Sep 24, 2019 at 22:11
• Well, yes, I assumed they are all measurable. Most of the time this type of convergence is defined for sequences of measurable functions. (especially when OP has the tag "measure theory" under the question). But yes, I should have written that earlier.
– Mark
Commented Sep 25, 2019 at 0:02
1. It means that $$\lim_{n\to \infty }f_n(x)=f(x)$$ for a measurable set $$E\subset X$$ such that $$\mu(X\setminus E)=0$$. That is: it doesn't converge just in a set of zero measure (also called a null set).

2. I see it just as a.e. pointwise convergence.

3. The almost everywhere concept is applied to many different things, not just convergence. Depending of the context it have different meanings. In this context it is a subtype of pointwise convergence.