Let f be integrable over E
If $E_n, n \in \mathbb{N}$ is a descending countable collection of measurable subsets of E, the $\int_{\bigcap_{n=1}^{\infty}E_n}f = lim_{n \to \infty} \int_{E_n} f$.
I got this for the case with union. Using the same idea, I tried to switch from the sets $E_n$ to a set$X_n = E_1$ \ $E_{n+1}$, for n $\geq$ 1 which is an ascending seq of measurable subsets whose union is $E_1$ \ $\bigcap_{n=1}^{\infty}E_n$. Then I want to apply the result for union, but I'm not sure how to go about this.
So $1_{E_{n}}f\rightarrow 1_{S}f$, where $S$ is such the intersection, now $1_{E_{n}}|f|\leq |f|$ and $|f|\in L^{1}$, the rest is just Lebesgue Dominated Convergence Theorem.
Indeed, if one can do it for union (increasing sequence of sets) then one can perform the trick that $F_{n}=E-E_{n}$, then $(F_{n})$ is increasing, then their union is $E-S$. Now observe that $\displaystyle\int_{E-E_{n}}=\int_{E}-\int_{E_{n}}$ and $\displaystyle\int_{E-S}=\int_{E}-\int_{S}$.