# Why do we need to define Lebesgue spaces using equivalence classes?

When we define an $L^{p}$ space for $1\leq p \leq \infty$, we say elements of this space are equivalence classes of functions which are equal almost everywhere and $$\int|f|^{p} dx < \infty$$

Why can we not say elements are functions which satisfy $\int|f|^{p} < \infty$ ?

I understand that if $g=f$ a.e. then $||f||_{L^{p}} = ||g||_{L^{p}}$ is this the reason for it?

EDIT :

The reason for asking is because I am studying an optimal control of PDEs course which says we need to be careful when considering the PDE :

$-\Delta y = f$ on $\Omega$

$y=0$ on $\partial \Omega$

...since we need to define what it means for $y=0$ on $\partial\Omega$, since $\partial\Omega$ has zero measure.

• Even more, if $g=f$ a.e. then $\|f-g\|_{L^p}=0$. – Aweygan Apr 11 '18 at 13:53
• Out of curiosity, how did you end up solving the issue of the boundary having 0 measure? – Kitegi Apr 15 '18 at 17:09
• Hi @Kitegi, since were dealing with functions in $L^{P}$ we can change their value on the boundary without changing the function. So we need a non-ambiguous definition of boundary value. We can write our functions y as limits of sequences of $y_{n}$ - functions continuous on the boundary of $\Omega$ then take our value of y on the boundary as the limit of $y_{n}$ on the boundary. For more infomation see the trace theorem. – Monty Apr 16 '18 at 17:03

In order for the $L^p$-norm to truly be a norm, it needs to be true that $\| f \|_{L^p} = 0 \implies f = 0$. But if $f$ is a measurable function and $\int |f|^p \, dx = 0$, we can only conclude that $f$ is zero almost everywhere.
• For many existence results based on functional analytic techniques it is enough to have a complete semi-normed space (i.e., $\|x\|=0$ may hold for $x\neq 0$). In this setting, only the uniqueness of limits is missing (but not the existence of limits of Cauchy sequences). – Jochen Apr 13 '18 at 12:05