Let
- $(\Omega,\mathcal A,\operatorname P)$ be a measure space
- $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra on $\Omega$
Since $\mathcal F\subseteq\mathcal A$, the identity $\operatorname{id}_\Omega$ on $\Omega$ is $\mathcal A$-$\mathcal F$-measurable and hence the $\mathcal F$-pushforward $\mu\circ\operatorname{id}_\Omega^{-1}$ of $\mu$ with respect to $\operatorname{id}_\Omega$ is well-defined. By definition $$\left(\mu\circ\operatorname{id}_\Omega^{-1}\right)(F)=\mu\left(\operatorname{id}_\Omega^{-1}(F)\right)=\mu(F)\;\;\;\text{for all }F\in\mathcal F\tag1$$ and hence $$\mu\circ\operatorname{id}_\Omega^{-1}=\left.\mu\right|_\mathcal F\;.\tag2$$ If $f\in\mathcal L^1\left(\left.\mu\right|_\mathcal F\right)$, then $f=f\circ\operatorname{id}_\Omega\in\mathcal L^1(\mu)$ and $$\int f\:{\rm d}\mu=\int f\circ\operatorname{id}_\Omega\:{\rm d}\mu=\int f\:{\rm d}\left(\mu\circ\operatorname{id}_\Omega^{-1}\right)=\int f\:{\rm d}\left.\mu\right|_\mathcal F\;.\tag3$$
So, $$\mathcal L^1\left(\left.\mu\right|_\mathcal F\right)\subseteq\left\{f\in\mathcal L^1(\mu):f\text{ is }\mathcal F\text{-measurable}\right\}\;.\tag4$$ Are we able to show that the other inclusion holds as well?
If we choose $(\Omega,\mathcal A):=\left(\mathbb N,2^{\mathbb N}\right)$, $\mu$ to be the counting measure on $(\Omega,\mathcal A)$ and $\mathcal F:=\left\{\emptyset,\mathbb N\right\}$, then $\mu$ is $\sigma$-finite while $\left.\mu\right|_\mathcal F$ obviously is not. This is not a counterexample, but it raises some doubts.
I've seen that some people use the notation $\mathcal L^1(\mathcal F,\mu)$ for the set on the right-hand side of $(4)$. This notation seems to be bad unless we've got equality in $(4)$.
Let $f$ be $\mathcal{F}$-measurable. You correctly point out the fact that if $f\in\mathcal L^1\left(\left.\mu\right|_\mathcal F\right)$ then $f\circ\operatorname{id}_\Omega\in\mathcal L^1(\mu)$. But the converse also holds.
In particular, if $f \in\mathcal L^1(\mu)$ and is $\mathcal{F}$-measurable, then $f\circ\operatorname{id}_\Omega\in\mathcal L^1(\mu)$ which means (applying my earlier remark) that $f\in\mathcal L^1\left(\left.\mu\right|_\mathcal F\right)$. This proves that $(4)$ is an equality.