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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.

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    $\begingroup$ Even more, if $g=f$ a.e. then $\|f-g\|_{L^p}=0$. $\endgroup$ – Aweygan Apr 11 '18 at 13:53
  • $\begingroup$ Out of curiosity, how did you end up solving the issue of the boundary having 0 measure? $\endgroup$ – Kitegi Apr 15 '18 at 17:09
  • $\begingroup$ 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. $\endgroup$ – Monty Apr 16 '18 at 17:03
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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.

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  • $\begingroup$ of course very simple, thanks!!! I have edited my question to include some background. $\endgroup$ – Monty Apr 11 '18 at 13:51
  • $\begingroup$ 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). $\endgroup$ – Jochen Apr 13 '18 at 12:05

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