# Confusion on "almost everywhere defined" function in $L^1$ space.

The following are excerpt from folland

This proposition shows that for the purposes of integration it makes no difference if we alter functions on null sets. Indeed, one can integrate functions $$f$$ that are only defined on a measurable set $$E$$ whose complement is null simply by defining $$f$$ to be zero (or anything else) on $$E^c$$. In this fashion we can treat $$\bar{\mathbb{R}}$$-valued functions that are finite a.e. as real-valued functions for the purposes of integration. With this in mind, we shall find it more convenient to redefine $$L^1(\mu)$$ to be the set of equivalence classes of a.e.-defined integrable functions on $$X$$, where $$f$$ and $$g$$ are considered equivalent iff $$f = g$$ a.e. This new $$L^1(\mu)$$ is still a complex vector space (under pointwise a.e. addition and scalar multiplication). Although we shall henceforth view $$L^1(\mu)$$ as a space of equivalence classes, we shall still employ the notation "$$f \in L^1(\mu)$$" to mean that $$f$$ is an a.e.-defined integrable function. This minor abuse of notation is commonly accepted and rarely causes any confusion.

I have the following questions:

1. What does it mean to be a.e. integrable? In other words, how do we know whether a.e defined functions are measurable and integrable? It is just whether the extension of this function with $$0$$s is integrable? Also Folland first said that "one can integrate functions defined on a measurable set $$E$$ whose complement is null by...". Functions defined a.e is not necessarily defined on a measurable set if the measure is not complete.
2. If two functions do not have the same domain, what does it mean by $$f=g$$ a.e? Is it whether the extension of $$f$$ and $$g$$ with $$0$$s are equal almost everywhere?
3. Precisely how do we treat $$\bar{\mathbb{R}}$$ functions that are finite a.e as real functions? Do we choose an arbitrary null set that contains all the infinities of the function and make them to be $$0$$? And since alter functions on null set does not makes no difference, it doesn't matter which null set we choose?
• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. Feb 12 at 13:36

A formal treatment for extending beyond real-valued functions is this: If $$(\Omega, \mathcal{F})$$ is any measurable space, then we can add a new element to create $$\tilde{\Omega} = \Omega \cup \{new_1\}$$ and we can create a new sigma algebra on $$\tilde{\Omega}$$: $$\tilde{\mathcal{F}}=\sigma(\mathcal{F}\cup \{new_1\})$$ that is only trivially modified. So we can easily get a new measurable space $$(\tilde{\Omega}, \tilde{\mathcal{F}})$$. We can add two or three new elements just as easily. The new elements could be \begin{align} \{new_1\}&=\{\infty\}\\ \{new_2\}&=\{-\infty\}\\ \{new_3\}&=\{DNE\} \mbox{ (DoesNotExist)} \end{align} So we can define \begin{align} \tilde{\mathbb{R}} &= \mathbb{R}\cup \{-\infty\}\cup\{\infty\}\\ \hat{\mathbb{R}} &= \mathbb{R} \cup \{-\infty\} \cup \{\infty\} \cup \{DNE\} \end{align} with corresponding sigma algebras.

So if we start with the usual Borel measurable space $$(\mathbb{R}, \mathcal{B}(\mathbb{R}))$$ and we want to add new points $$\{\infty\}$$ and $$\{-\infty\}$$, we can use $$\tilde{\mathbb{R}}$$ and the sigma algebra, call it $$\mathcal{B}(\tilde{\mathbb{R}})$$, will only be trivially modified. We could do the same with $$\hat{\mathbb{R}}$$. Your excerpt of Folland does not define $$\overline{\mathbb{R}}$$, so I do not know if it is the same as my $$\tilde{\mathbb{R}}$$ or my $$\hat{\mathbb{R}}$$ (or something else). Without making any assumptions, I will use my $$\tilde{\mathbb{R}}$$ in the following.

1. It is indeed true that a function $$f:X\rightarrow\tilde{\mathbb{R}}$$ must be assigned a value for all $$x \in X$$ (we are allowed to use $$f(x)=\infty$$ or $$f(x)=-\infty$$).

2. For measurable functions $$f:X\rightarrow\tilde{\mathbb{R}}$$ and $$g:X\rightarrow\tilde{\mathbb{R}}$$, we say $$f=g$$ a.e. if $$\mu(\{x \in X: f(x)\neq g(x)\})=0$$.

1. If a measurable function $$f:X\rightarrow\tilde{\mathbb{R}}$$ is "a.e.-finite" then $$\mu(\{x \in X: f(x) \notin \mathbb{R}\}) = 0$$ The theory of integration works just as well for measurable real-valued functions and measurable a.e.-finite functions.

Also 1: It can be shown that if $$f:X\rightarrow\tilde{\mathbb{R}}$$ is measurable, then for any set $$E^c\subseteq X$$ such that $$\mu(E^c)=0$$, we can change $$f(x)$$ to anything we like for $$x \in E^c$$ and (provided that our change preserves measurability) that does not change $$\int f$$. It can be shown that making $$f(x)=0$$ for all $$x \in E^c$$ preserves measurability (and does not change $$\int f$$). It can be shown that making $$f(x)=\infty$$ for all $$x \in E^c$$ preserves measurability (and does not change $$\int f$$). As you observe in your question, it is indeed sometimes possible to change $$f(x)$$ for $$x \in E^c$$ to make $$f$$ a nonmeasurable function (so don't do that).

Also 2: Some people would say the function $$f:X\rightarrow\tilde{\mathbb{R}}$$ has domain $$X$$, others would say it has domain $$\{x \in X: f(x) \in \mathbb{R}\}$$. So to avoid confusion I have not used the word "domain" anywhere in my answer.

Also 3: Some people would say summations $$\sum_{i=1}^{\infty}1/i$$ are "not defined" while others would say they are "infinity." So the conventions behind "not defined" and its relation to "not finite" are not fixed, which may add to general confusion around this topic. This is why it is useful to formally define the sets $$\tilde{\mathbb{R}}$$ and $$\hat{\mathbb{R}}$$ if they are needed.

Also 4: An advantage of formally using measurable functions $$f:X\rightarrow \mathbb{R}\cup \{\infty\}\cup\{-\infty\}\cup\{DNE\}$$, with the sigma algebra $$\sigma(\mathcal{B}(\mathbb{R})\cup\{\infty\}\cup\{-\infty\}\cup\{DNE\})$$ is that it allows the function $$f$$ to take the value DNE on arbitrary measurable subsets of $$X$$ (including subsets of positive measure).

• I replaced the downvote with an upvote. I will remove my comments as well. Thank you.
– user1266745
Feb 12 at 12:20
• @jwhite : Thanks for your help! Feb 13 at 0:33