real analysis - Integral on a completion of measure space

Let $(X,{\cal M}, \mu)$ be a measure space with completion $(X,\bar{{\cal M}},\bar{\mu})$. If $f \in {\cal L}^1(X,{\cal M},\mu)$, show that $f \in {\cal L^1}(X,\bar{\cal M},\bar{\mu})$ and $\int fd\mu = \int f d\bar{ \mu}$

My attempt: firstly to show that $f$ is $\bar {\cal M}$ measurable, let $E \in {\cal B}_{\mathbb{C}}$, then $f^{-1}(E) \in \cal{M}$, by the definition of completion, $\bar{\cal M} := \{F \cup G: F \in {\cal M}, G \subseteq N, N \in {\cal M}, \mu(N) = 0\}$, if we set $G = \emptyset$, then we can have $f^{-1}(E) \in \bar {\cal M}$, hence $f$ is $\bar {\cal M}$ measurable.

Furthermore, $f \in {\cal L}^1(X, \bar{\cal M},\bar{\mu})$ will directly follow from $\int f d\mu = \int f d \bar{\mu}$, but how to show this? I appreciate your help!

Answer

It is enough to prove it for nonnegative $f$.

Let $S$ be the set of finite sums $\sum_{k=1}^na_n\mu(A_k)$ where the sets $A_k$ are $\mathcal M$-measurable, the $a_k$ are nonnegative, and finally $\sum_{k=1}^na_n1_{A_k}\leq f$.

Let $\bar S$ be the set of finite sums $\sum_{k=1}^na_n\mu(A_k)$ where the sets $A_k$ are $\bar{\mathcal M}$-measurable, the $a_k$ are nonnegative, and finally $\sum_{k=1}^na_n1_{A_k}\leq f$.

Then $\int fd\mu=\sup S$ and $\int fd\bar{\mu}=\sup\bar S$.

It is evident that $S\subseteq\bar S$ so $\int fd\mu\leq\int fd\bar{\mu}$.

For every $s=\sum_{k=1}^na_n\mu(A_k)\in\bar S$ we can find sets $B_k\in\mathcal M$ such that $B_k\subseteq A_k$ and $\mu B_k=\bar{\mu} A_k$. Then $s=\sum_{k=1}^na_n\mu(A_k)=\sum_{k=1}^na_n\mu(B_k)\in S$.

So we also have $\bar S\subseteq S$ so that $\int fd\bar{\mu}\leq\int fd\mu$.