Some questions regarding the $L^2$ space

Question

Let $(\Xi,\boldsymbol{\Xi},\mu)$ be a measure space, and let $(\mathbb{R},\mathcal{B}_\mathbb{R})$ be the Borel space associated to $\mathbb{R}$. Also, let $$\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})=\bigg\{f:(\Xi,\boldsymbol{\Xi})\to(\mathbb{R},\mathcal{B}_\mathbb{R})\,\,\Big|\,\int f^2\,\mathrm{d}\mu<\infty\bigg\}$$ be the set of all (real-valued) square-integrable (measurable) functions with respect to (the measure) $\mu$.

Then can we say the following?

1. When the Lebesgue space $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ is viewed as a set, it is defined as the quotient set: $$L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})_\text{set}:=\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\,\,/\sim,$$ where $\sim$ is an equivalence relation on $\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ such that: $$f\sim g\quad:\Leftrightarrow\quad f-g=0\,\text{ $\mu$-almost everywhere.}$$ Is this the definition of $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ as a set?

2. If so, what does '$\mu$-almost everywhere' actually mean in the above definition? Some authors seem to imply that (but of course I may be wrong with this interpretation): $$f(\xi)\neq g(\xi)\quad\Rightarrow\quad \mu(\xi)=0$$ when they say that if $f=g$ then $f$ and $g$ differ only on a set of measure zero. Here $\xi\in\Xi$. However, for this to work, we require that $\{\xi\}$ is an element of the $\sigma$-algebra $\boldsymbol{\Xi}$ to make sure that $\{\xi\}\mapsto\mu(\xi)$ is well-defined. This therefore imposes a hidden requirement on the definition of $\boldsymbol{\Xi}$ (which I don't like). Or is this the correct interpretation? $$f-g=0\,\text{ $\mu$-almost everywhere}\quad\Leftrightarrow\quad\int (f-g)^2\,\mathrm{d}\mu=0.$$

3. The reason why we want to define such an equivalence relation $\sim$ on $\mathcal{L}^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ is that if we equip $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ with the bilinear map: $$\langle\,\cdot\,,\cdot\,\rangle:L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\times L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\to\mathbb{R}\quad:\Leftrightarrow\quad\langle f,g\rangle=\int fg\,\mathrm{d}\mu,$$ then $\langle\,\cdot\,,\cdot\,\rangle$ qualifies as an inner product on $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ because it satisfies the positive-definite condition of an inner product. Is this the correct motivation?

4. $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ is by definition a vector space over $\mathbb{R}$, and no further structure is assumed on $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})_\text{set}$.

5. If nothing is said, all (equivalence classes of) measurable functions in $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ are always assumed to be maps from $(\Xi,\boldsymbol{\Xi})$ into the Borel measurable space $(\mathbb{R},\mathcal{B}_\mathbb{R})$, aren't they?

(Disclaimer: I am not a mathematician but an engineer studying functional analysis; thus, the notation used here may be non-standard or simply incorrect. If that is the case, please let me know. I would appreciate it very much.)

Thank you, Frederick.

Answer 1

Concerning (2), we say that $f = g$ $\mu$-a.e. (or $f - g = 0$ $\mu$-a.e.) if $\mu(\Theta \setminus \{f - g = 0\}) = 0$. This makes sense as long as $f$ and $g$ are measurable functions from $(\Theta,\mathbf{\Theta})$ into $(\mathbb{R},\mathscr{B}_{\mathbb{R}})$ since then $\{f - g = 0\} \in \mathbf{\Theta}$.

It turns out that this is, in fact, equivalent to the condition $\int_{\Theta} (f - g)^{2} \, d \mu = 0$. However, this requires that we define the Lebesgue integral, whereas the previous definition I gave is a "primitive notion."

The answer to (3) is "yes." However, it's a bit like saying "We define $\mathbb{R}$ because $\mathbb{Q}$ isn't a complete metric space." That may technically contain some kernel of truth, but there are "philosophical" or "emotional" reasons why we would like to work with $\mathbb{R}$ in addition to $\mathbb{Q}$ (i.e. we might like to believe that a "continuum" or a "line" has some physical significance). Similarly, in the settings in which measure theory is relevant, it tends to be natural to think of two functions that agree $\mu$-almost everywhere to be the same. ($\mu$ sees them the same way, let's say; or we added "some dust" to one of the functions, which is irrelevant.)

Concerning (5), yes, technically, an element of $L^{2}(\Theta,\mathbf{\Theta},\mu;\mathbb{R})$ is an equivalence class of measurable functions mapping $(\Theta,\mathbf{\Theta})$ into $(\mathbb{R},\mathscr{B}_{\mathbb{R}})$. The one caveat is nothing is lost or gained if we replace $(\Theta,\mathbf{\Theta})$ by $(\Theta,\mathbf{\Theta}_{\mu})$, where $\mathbf{\Theta}_{\mu}$ is the completion of $\mu$ with respect to $\mathbf{\Theta}$. Hence, formally, it might be better to incorporate completeness of $(\Theta,\mathbf{\Theta},\mu)$ into the definition of $L^{2}(\Theta,\mathbf{\Theta},\mu;\mathbb{R})$ at the outset. Someone can correct me if I'm wrong, but I'm not aware of any instances where this matters. (It's a bit like the space $\mathcal{L}(V)$ of continuous linear operators on a normed space $V$. $\mathcal{L}(V)$ is unchanged if $V$ is replaced by its completion.)

Answer 2

1. I am assuming that by ${\cal L}^2$ you mean the square integrable functions and by $L^2$ you mean the equivalence classes. Both of these are sets and you have $L^2 = {\cal L}^2 / \sim$. (Both are vector spaces with the appropriate definitions of $+,\cdot$.)

2. $f=g $ $[\mu]$ ae. means there is a set of measure zero (null set) $N$ such that $f(x)=g(x)$ for all $x \notin N$.

3. The reason for the equivalence classes is so that $f \mapsto \sqrt{\int |f|^2d \mu}$ is a norm rather than a seminorm.

4. $L^2$ is a set, I am not sure how you distinguish the two.

5. No. An element of $L^2$ is the equivalence class of (square integrable) measureable functions $\Xi \to \mathbb{R}$.

In many situations one can be sloppy about the distinction, just as we rarely think of real numbers as equivalence classes. Sometimes one needs to be careful, for example, to define the $\sup$ norm (for $L^\infty$) of support of an equivalence class.