# Why is it nonsensical to evaluate $f(0)$ for $f\in L^p$

(Lieb and Loss’ Analysis, pg. 41-43)

“Let $\Omega$ be a measurable space with (positive) measure $\mu$ and let $1\le p<\infty$ We define $L^p(\Omega,d\mu)$ to be the following class of functions $$L^p(\Omega,d\mu)=\{f\mid f:\Omega\to\mathbb{C}, \ f \ \text{is} \ \mu\text{-measurable}, \ |f|^p \ \text{is summable}\}$$ $\dots$ redefine $L^p(\Omega, d\mu)$ so that its elements are not functions but equivalence classes of functions $\dots$ space that should be called $L^p\dots$ we note that it makes no sense to ask for the value $f(0)$, say, of an $L^p$-function.

Why is this last line, the italicized text, true? An equivalence class of functions is a set of functions that agree almost everywhere nonetheless, it is a set of functions. So therefore, as long as zero is an element of $\Omega$, it is valid to evaluate it there.

• $L^p$ consists of classes of functions, not functions: they are functions modulo functions that vanish almost everywhere. In particular, if $f$ and $g$ differ in one value, they define the same class. – Pedro Tamaroff Feb 9 '18 at 23:32
• $L^p$ is often defined as set of functions, but in any metric space setting, we think of $L^p$ as set of equivalence classes – zhw. Feb 9 '18 at 23:34

And let's be more formal than usual - say $[f]$ is the class of all $g$ with $f=g$ almost everywhere.
Then $f$ is not an element of $L^p$, strictly speaking it's $[f]$ that's an element of $L^p$. The only sensible definition of $[f](0)$ would be $$[f](0)=f(0).$$But that's not well-defined: If we choose $g$ so $f=g$ almost everywhere but $f(0)\ne g(0)$ then $[f]=[g]$ but that notation would imply $[f](0)\ne[g](0)$.
Consider the usual Lebesgue measure in Euclidean. Now let $f(x)=0$ for all $x$ and $g(x)=\chi_{\{0\}}(x)$, so $g(0)=1$ and $g(x)=0$ for all $x\ne 0$, but $[f]=[g]$.
Here's a fun one. Consider the function \begin{align} f(x) = \begin{cases} \sin \frac{1}{x}& \ \text{ if } x\neq 0\\ c& \ \text{ if } x = 0 \end{cases} \end{align} on $[0, 1]$, where $c$ can be anything. Observe that $f\in L^p[0, 1]$ for any $1\le p\leq \infty$.