What is the relationship between $L_{P}(0,1)$ and $L_{P}[0,1]$? $L_{P}[0,1]$ be the set of measurable functions  $f : [0,1]\rightarrow R$ such that
$\int |f(x)|^{p} dx<\infty$. What is the relationship between $L_{P}(0,1)$  and $L_{P}[0,1]$?
 A: If $L_p(0,1)$ means the set of measurable functions $f:(0,1)\to R$ with $\int|f(x)|^p dx < \infty$,
then they are the 'same' up to equality almost everywhere. Formally, you need to embed $L_p(0,1)$ into $L_p[0,1]$ by extending it's elements onto $[0,1]$.
A: Assuming you mean the normal(Lebesgue) measure, then noting that the endpoints are sets of measure zero, we can thow them out of the integral and not change its value.  Note that in $L^p[0,1]$, if we fix a function $f$ on $(0,1)$ and then assign arbitrary values to their endpoints, then they are the same element in $L^p$, as it is actually a space of equivalence classes of functions.
In the more general setting, if we take any measure space $(\Omega, \Sigma, \mu)$ and a set $A$ of measure zero.  Then 
\begin{align}
\int_\Omega f d\mu = \int_{\Omega\backslash A} f d\mu + \int_A f d\mu =  
\int_{\Omega\backslash A} f d\mu
\end{align}
So $L^p(\Omega) = L^p(\Omega\backslash A)$.  Note that the this also works for $L^\infty$, as it uses the "essential supremum" which allows us to change a set of measure zero.
