How do you rigorously explain the fact that $u \in L^p$ can be non defined over sets of measure 0?

Question

In all the definitions of $L^p(\Omega)$ spaces I have been given these are defined to be the set of functions $f: \Omega \to \mathbb{R}$ whose norm $||\cdot||_{L^p}$ is finite. We define is as the quotient with the equivalence relation $f=g$ if and only if $f=g$ almost everywhere.

Now, the books I am dealing with say that $f$ can be non-defined on sets of measure $0$. But the definition of function explicitly says $f(x)$ is defined on every $x \in \Omega$.

Do we really relax this last condition to be "defined almost everywhere" or the books I am dealing with want to express that we cannot make sure what the value at a especific set of measure $0$ is as it can always be redefined (but it is defined)?

Answer 1

It doesn't matter - the two versions of the definition give isometrically isomorphic spaces. Allowing functions to be undefined on a set of measure zero can be convenient, for example allowing us to refer to $f(x)=|x|^{-1/2}$ as an element of $L^1([-1,1])$ without having to define $f(0)$. Or allowing us to define $f=\lim f_n$ when the limit only exists almost everywhere, etc.

It's so clear that it doesn't matter that people do use the second version, or write as though they were using it, without every worrying about giving a precise statement of the second version of the definition. If I wanted to state that definition precisely I'd probably start like so:

If $\mu$ is a measure on $X$ then an almost function on $X$ is a function $f:X\setminus E\to\Bbb C$ for some set $E$ with $\mu(E)=0$.

Then if you feel like it you can give definitions of the sum and product of two almost functions, the integral of an almost function, what it means for two almost functions to be equal almost everywhere, etc, finally defining $L^p$ as a certain space of equivalence classes of almost functions.

Answer 2

Note that $L^p(\Omega)$ is not a set of functions, but a set of equivalence classes of functions; specifically, we define an equivalence relation by saying for functions $f, g : \Omega \rightarrow \mathbb{R}$, we have $f \sim g$ if and only if the set $\{ \omega \in \Omega | f(\omega) \not= g(\omega) \}$ has measure 0 (it can be checked that this does indeed define an equivalence relation). You've somewhat alluded to this is your first paragraph.

Saying that "$f \in L^p(\Omega)$ can be undefined on a set of measure $0$" is a sloppy way of saying that given an element of $L^p(\Omega)$ (that is, an equivalence class of functions under $\sim$), the corresponding functions in the equivalence class may disagree on any set of measure $0$. In fact, given an element of $L^p(\Omega)$, it doesn't make sense to talk about its value at any element of $\Omega$ (if the measure space $\Omega$ has no atoms), since given any point $\omega \in \Omega$, we can find two functions in our equivalence class that differ at that point.

It's also worth bearing in mind that when speaking about $L^p$ spaces, people are often a little sloppy and might say things like "for a finite measure space $\Omega$, the set $C_b(\Omega)$ of bounded continuous functions is a subset of $L^p(\Omega)$". Strictly speaking, this doesn't make sense, since $C_b(\Omega)$ is a set consisting of functions, whilst $L^p(\Omega)$ is a set consisting of equivalence classes of functions, so it's not possible for one to be a subset of the other. What is understood by this remark is "the set of equivalence classes $\{ [f] | f \in C_b(\Omega) \}$ is a subset of $L^p(\Omega)$" - this is a bit more of a mouthful, and generally speaking it is understood that the first remark is really shorthand for the second (in the same way that the distinction between integers and equivalence classes of integers is often implicit when doing modular arithmetic).