What does mean : element of $L^p$ are equivalence class rather than function.

Question

What does mean : An element of $L^p$ is rather an equivalent class that a function ? If $f\in L^p$, why don't we see it as a function (it's always what I did until now, but why is it not totally correct ?) What is the subtlety with these "equivalent class" ?

Answer 1

Let's say you have a measurable function $f \colon X \rightarrow \mathbb{R}$ on $(X,\mu,\Sigma)$. You can say two distinct things about $f$:

1. The function $f$ is $p$-integrable meaning $\int_X |f|^p d\mu < \infty$. Let's denote the space of $p$-integrable functions on $X$ by $L^p(X,\mu, \Sigma)$.
2. The function $f$ "belongs to $L^p$". This is actually an abuse of terminology since the space $L^p$ is not a space of functions but a space of equivalence classes $$ \mathcal{L}^p(X,\mu,\Sigma) := L^p(X,\mu,\Sigma) / \sim $$ where $f \sim g$ if and only if $f - g = 0$ a.e. That is, when you say that $f$ belongs to $\mathcal{L}^p$, you actually mean that $[f] \in \mathcal{L}^p(X,\mu,\Sigma)$.

The reason the $\mathcal{L}^p$ spaces are defined as a quotient of actual $p$-integrable functions $L^p$ by an equivalence relation and not just as the spaces of $p$-integrable functions is that you want to turn them into normed vector spaces using the norm

$$ \| f \|_p = \left(\int_X |f|^p \, du\right)^{1/p}. $$

This is not a norm on $L^p$ but only a semi-norm since it is possible that $\| f \|_p = 0$ even though $f \neq 0$. By taking the quotient, the semi-norm descends to an honest-to-god norm on the quotient space $\mathcal{L}^p$.

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Thus, an element $f \in L^p(X,\mu,\Sigma)$ is an actual function on $X$ so you can talk about (say) the value of $f$ at a point $x \in X$. However, an element $[f] \in \mathcal{L}^p(X,\mu,\Sigma)$ is not a function on $X$ but an equivalence class of functions on $X$ and so (say) the value of $[f]$ at a point $x \in X$ doesn't make sense (since it is possible that $[f] = [g]$ but $f(x) \neq g(x)$ so the operation of evaluating an equivalence class at a point $x \in X$ is not well-defined).

Answer 2

This kind of thing occurs frequently in mathematics.

You may say a "rational number" $\frac{7}{12}$ is not the pair $(7,12)$, but rather an equivalence class of such pairs, so that $\frac{7}{12}=\frac{-7}{-12}=\frac{21}{36}=\dots$.

You may say that a "real number" is an equivalence class of Cauchy sequences of rational numbers, where we allow two different Cauchy sequences to represent the same real number. Agreed, this can be confusing, judging from all he questons about whether $0.\overline{9} = 1$ or not.

You may say that $\mathbb C = \mathbb R [X]/(X^2+1)$, so that a complex numbr is an equivalence class of polynomials over the reals. For beginners, we may disguise this by saying: "A complex number is of the form $a+bi$ and for computations use the ordinary rules together with $i^2=-1$".

Answer 3

Let $(X,{\cal A},\mu)$ be a measure space, and let $\mu^*$ be the completion of $\mu.$ IF $f,g\colon X\longrightarrow\mathbb R$ are measurable functions, we usually identify $f$ with $g$ when the set $\{x\in X\,|\,f(x)\neq g(x)\}$ has null $\mu^*$-measure. This identification is actually an equivalence relation. Thanks to it, it is true that$$\lVert f-g\rVert_p=0\iff f\sim g,$$where $\sim$ is the equivalence relation that I have described. Otherwise, we could have distinct functions such that the distance between them would be $0$. That is, $\bigl(L^p(\mu),\lVert\cdot\rVert_p\bigr)$ would not be a normed-vector space.