# $\lim_ {n \to \infty} f_n$ is $\Sigma$-measurable

Let $(X,\Sigma)$ be a measurable space and let $\{f_n\}_{n \in \mathbb{N}}$ be a sequence of $\Sigma$-measurable real valued functions with domains included in $X$.

What does it mean to say "domains included in $X$"? Why is this significant?

We define a function $\lim\limits_{n \to \infty} f_n$ by writing

$$\lim\limits_{n \to \infty} (f_n)(x) = \lim\limits_{n \to \infty} f_n (x)$$

for $x \in \cup_{n \in \mathbb{N}} \cap_{m \geq n} \text{dom } f_m$ for which the limit exists in $\mathbb{R}$.

Then $\lim\limits_{n \to \infty} f_n$ is $\Sigma$-measurable.

What does $x \in \cup_{n \in \mathbb{N}} \cap_{m \geq n}\text{dom } f_m$ mean? How can it be interpreted?

• For "domains included in $X$", as far as I can see this is redundant since $f_n^{-1} (\mathbb{R}) \in \Sigma$ and surely $f_n^{-1}(\mathbb{R})\subset X$ – user160738 Dec 19 '15 at 16:10
• There is a special command to typeset the limit: \lim_{ ... }. Also \sum gives a large $\sum$, while \Sigma gives a normal sized $\Sigma$. – Winther Dec 19 '15 at 17:24

$$\cup_{n \in \mathbb{N}} \cap_{m \geq n}\hbox{dom}f_m$$
consists of precisely those elements of $X$ where $f_m(x)$ is well-defined from a certain value of $m$ onwards. That is the condition you need to impose on $x$ to be able to even define the concept of $\lim_{n\to\infty}f_n(x).$