Lp space over measure space and over its completion is the same. It is exercise of Tao's 1.3.1

Let $(X, \chi,\mu)$ and $(X, \overline{\chi}, \mu)$ is a measure space, where $\overline{\chi}$ is the completion of $\chi$. Show that $L^{p}(X, \chi, \mu)$ is isomorphic to $L^{p}(X, \overline{\chi}, \mu)$ using the obvious candidate for the isomorphism. Be- cause of this, when dealing with Lp spaces, we will usually not be too concerned with whether the underlying measure space is complete.

Actually I think this is obvious; but my answer is this.

Let $\phi: L^{p}(X, \chi, \mu) \to L^{p}(X, \overline{\chi}, \mu)$ as $f \mapsto f$.Since $\overline{\chi}$ is topology adding $\sigma-algebra$ of $\mu-$measurable zero set, $f \in L^{p}(X, \chi, \mu)$ is also absolutely integrable on $(X, \overline{\chi}, \mu)$. 
  Edited part start For showing subjectivity, think about nonempty measurable set $E \in \overline{\chi}$. Then $E$ has $E_{1}, E_{2} \in \chi$ such that $E_{1} \subseteq E_{2} \subseteq E$ and $\overline{\mu}(E_{1}\setminus E_{2}) = 0$ and $\overline{\mu}(E) = \mu(E_{1}) = \mu(E_2)$. Hence for any $\overline{\mu}$ measurable $f = \sum_{i\in A}a_{n}1_{\overline{E_{i}}}$, construct $g = \sum_{i\in A}a_{n}1_{E_{i}}$ where $\overline{E_{i}} \in \overline{\chi}$, $E_{i} \in \chi$ and $\overline{\mu}(\overline{E_{i}}) = \mu(E_i)$. Then for any $f \in L^{p}(X, \overline{\chi}, \mu)$, there exist $g \in (X, \chi, \mu)$ such that $g \in [f]_{\overline{\chi}}$

Do you guys think this answer is not rigorous or weird? Please let me know. I'm not sure whether this answer is right or wrong.
 A: Define $i:L^p(X,\chi,\mu) \to L^p(X,\bar{\chi},\bar{\mu})$ by
$i([f]_{\chi}) = [f]_{\bar{\chi}}$. That is, if $[f]_{\chi}$ is an equivalence class in $L^p(X,\chi,\mu)$, then $i([f]_{\chi})$ is the equivalence class with respect to $\bar{\chi}$, in particular we have $[f]_{\chi} \subset [f]_{\bar{\chi}}$, and so it is clear that $i$ is injective.
To see that $i$ is well defined, note that if $[f]_{{\chi}} \in L^p(X,\chi,\mu)$, then $f$ is $\mu$-measurable and $\int |f|^p d \mu < \infty$. It is straightforward to see that $f$ is $\bar{\mu}$-measurable, and $\int |f|^p d \bar{\mu} < \infty$, and so $i([f]_{\chi}) = [f]_{\bar{\chi}} \in L^p(X,\bar{\chi},\bar{\mu})$.
To finish, we need to show that $i$ is surjective. Suppose $[f]_{\bar{\chi}} \in L^p(X,\bar{\chi},\bar{\mu})$. Then $f$ is $\bar{\mu}$-measurable, and so there exists a sequence of $\bar{\mu}$-measurable simple functions $\bar{s}_n$ such that $|\bar{s}_n(x)| \le |f(x)|$ , and $\bar{s}_n(x) \to f(x)$ for all $x \in X$.
Each $\bar{s}_n$ has the form $\bar{s}_n = \sum_k \alpha_k 1_{\bar{A_k}}$, where the $\bar{A_k}$ are $\bar{\mu}$-measurable. Since this is the completion measure, there are $\mu$-measurable sets $A_k \subset \bar{A_k}$, with $\bar{\mu} (\bar{A_k} \setminus A_k) = 0$. Let $s_n = \sum_k \alpha_k 1_{A_k}$, and note that $s_n$ is $\mu$-measurable, and $s_n(x) = \bar{s}_n(x)$ for ae. [$\bar{\mu}$] $x$ (note the measure is $\bar{\mu}$). Let $D = \cup_n \{ x | s_n(x) \ne \bar{s}_n(x) \}$, and note that $\bar{\mu} D = 0$.
Since $\bar{s}_n(x) \to f(x)$ for all $x \in X$, we have $s_n(x) \to f(x)$ for ae. [${\mu}$] $x$. To see this, Let $E \subset X$ be the set for which $s_n$ does not converge to $f$, and suppose $\mu E >0$. Then $\mu E = \bar{\mu} E =  \bar{\mu} (E \setminus D) >0$, which contradicts $\bar{s}_n(x) \to f(x)$ for all $x \in X$. Hence $\mu E = 0$.
Finally, let $t_n$ be the simple functions $t_n = s_n \cdot 1_{E^c}$, and  note that the $t_n$ are $\mu$-measurable and $t_n(x) \to f(x) 1_{E^c}(x)$ for all $x$, hence $g=f \cdot 1_{E^c}$ is $\mu$-measurable. Since $\mu E = 0$, we have
$i([g]_{\chi}) = [g]_{\bar{\chi}} = [f]_{\bar{\chi}}$, hence $i$ is surjective.
(To summarize, if $f$ is $\bar{\mu}$-measurable, then there is a $\mu$-measurable $g$ such that $g \in [f]_{\bar{\chi}}$.)
