Let $\mathcal{G} \subseteq \mathcal{F}$ and $X: \Omega \to \mathbb{R}$ be a $\mathcal{G}-$measurable random variable show that:
$X \in L^{1}(P) \iff X \in L^{1}(P\vert_{\mathcal{G}})$
and in this case $E_{P}[X]=E_{P\vert_{\mathcal{G}}}[X]$
My idea: Although the statement seems very intuitive, I am having a tough time proving it.
Since $X$ is $\mathcal{G}-$measurable we know that $X^{-1}(A)\in \mathcal{G}$ for any $A \in \mathcal{B}(\mathbb R)$ and hence we can partition $\Omega$ into some disjoint $(A_{i})_{ \{i=1,...,n\} }\subseteq \mathcal{G}$:
Hence $E_{P}[\vert X\vert]=\sum\limits_{i=1}^{n}\int\limits_{A_{i}}\vert X\vert dP=\sum\limits_{i=1}^{n}\int\limits_{A_{i}}\vert X\vert dP\vert_{\mathcal{G}}=E_{P\vert_{\mathcal{G}}}[\vert X\vert]$
This is the only solution I could think of, but I do not believe it is correct.
Well, you just have to ask yourself, what's the $L^1$ norm of $X$, and how does it arise? Well, since $|X|$ is $\mathcal{G}$-measurable, there exists a sequence, $(u_n)_{n\in\mathbb{N}}$ of $\mathcal{G}$-measurable simple functions increasing pointwise to $|X|$.
By monotone convergence, for any measure $\mu$ on either $\mathcal{G}$ or $\mathcal{F}$, we now have
$$ \int |X| \textrm{d} \mu=\lim_{n\to\infty} \int u_n\textrm{d}\mu, $$ so it suffices to prove the result for simple, positive, $\mathcal{G}$-measurable functions. But this follows immediately from the definition of the restricted measure.