Doubt on property of norm on $L_1(\mathbb{R})$

Question

we define the space of absolutely integrable functions $$L_1\left( \mathbb{R} \right):=\left \{ f:\mathbb{R} \rightarrow \mathbb{C}~ \bigg{|}~\int_{-\infty}^{\infty}|f(x)|\mathrm{d}x <+\infty\right\}$$ We define the norm on $L_1 \left( \mathbb{R} \right)$ as $$\| f\|_1:=\int_{-\infty}^{\infty}|f(x)|\mathrm{d}x$$ My doubt is that is this really a valid norm ? As this does not satisfy the property that $\|f\|=0 \iff f=0$. Its clearly satisfies the reverse direction, but not the forward direction. We can consider an example $$\tilde{f}(x): = \left\{\begin{aligned}1 ~~&\text{if}~~x=0\\0 ~~&\text{if}~~x\in \mathbb{R}\setminus\{0\}\end{aligned}\right.$$ This satisfies $\|\tilde{f}\|=0$ even if $f$ is not entirely the zero function.

My instructor said this has something to do with "measure theory" which i'm not much familiar with. I don't really $\textit{grok}$ this over the top explanation. Will appreciate any help understanding this concept.

Answer 1

The Banach space $L^1(\Bbb R)$ is not a set of functions, but rather a set of equivalence classes of functions. Specifically, $L^1(\Bbb R)$ is the quotient set of the set of measurable and Lebesgue integrable functions $\Bbb R\to \Bbb R$ with respect to the equivalence relation $$f\sim g\iff f=g\text{ almost everywhere}$$ Therefore your $\widetilde f$, or rather the class $\left[\widetilde f\right]$ of your function, is actually the zero element of $L^1(\Bbb R)$.

Added: I cannot quite explain everything about measure theory in this space (fat chance I could do it in any space, actually): if you want to read into it, Royden's Real Analysis contains a couple of good introductory chapters, at least for the purpose of understanding Lebesgue integral, $L^p$ spaces and such. I can surmise the "almost everywhere" part.

A measure space is the given of a triple $(X,\mathcal E,\mu)$, where:

• $X$ is a set;
• $\mathcal E$ is a $\sigma$-algebra of sets on $X$, id est a family of subsets of $X$ such that:
  1. $X\in \mathcal E$,
  2. for all $A,B\in \mathcal E$, $B\setminus A\in\mathcal E$,
  3. for all sequences $\{A_n\}_{n\in\Bbb N}$ of elements of $\mathcal E$, $\bigcup_{n\in\Bbb N}A_n\in\mathcal E$;
• $\mu$ is a (positive and $\sigma$-additive) measure on $(X,\mathcal E)$, id est a function $\mu:\mathcal E\to[0,\infty]$ such that $\mu(\emptyset)=0$ and, for all sequences $\{A_n\}_{n\in\Bbb N}$ of disjoint elements of $\mathcal E$, $\mu\left(\bigcup_{n\in\Bbb B}A_n\right)=\sum_{n\in\Bbb N}\mu(A_n)$.

An interesting instance of such a thing is $X=\Bbb R^n$, $\mathcal E=\text{Lebesgue-measurable sets}$ and $\mu$ Lebesgue measure. In this setup, a property $P$ of elements of $X$ is said to hold almost everywhere if the set of the elements that fails it is contained in a subset of measure zero. In other words if there is some $N\in\mathcal E$ such that $\mu(N)=0$ and $\{x\in X\,:\, \neg P(x)\}\subseteq N$. By extension, $f=g$ almost everywhere means that $f(x)=g(x)$ holds almost everywhere.