Suppose I have a set $E\subset \mathbb{R}^n$ with $E=\bigcup_k E_k$ a countable (almost disjoint) union of open sets with positive Lebesgue measure.
I know for $f\in L^1(E)$ we have that $\int_{E}f $$d\mu=\sum_k \int_{E_k} f d\mu$. The proof is standard. Does this result holds in general $L^p$?
We can make the substitution $g=|f|^p$ to obtain $\int_{E}g $$d\mu=\sum_k \int_{E_k} g d\mu$. Now we just raise everything to $1/p$ and obtain
$(\int_{E}g $$d\mu)^{1/p}=(\sum_k \int_{E_k} g d\mu)^{1/p}$.
Now back-substitute $|f|^p$ and we obtain $(\int_{E}|f|^p $$d\mu)^{1/p}=(\sum_k \int_{E_k} |f|^p d\mu)^{1/p}$.
Now let's raise everything to the $p$ power and we have
$(\int_{E}|f|^p $$d\mu)=(\sum_k \int_{E_k} |f|^p d\mu)$. This is equivalent to
$||f||_{L^p(E)}^{p}=\sum_k ||f||_{L^p(E_k)}^p$
So the original result does not hold, but the modified result does.
You cannot expect equality: Take $f(x) = \exp(-x)$ on $[1,\infty)$, $p=2$ and $E_n=[n,n+1)$. Then we have $$M_n:=\int_{n}^{n+1} \exp(-2x) \, \mathrm{d} x = \frac{\exp(-2n) - \exp(-2(n+1))}{2}$$ and $$\Bigg( \sum_{n=1}^\infty M_n \Bigg)^{1/2} = \Bigg( \int_1^\infty \exp(-2x) \, \mathrm{d} x \Bigg)^{1/2} = \frac{e^{-1}}{\sqrt{2}},$$ but by the geometric-series $$ \sum_{n=1}^\infty M_n^{1/2} = \frac{e^{-1}}{\sqrt{2}} \frac{\sqrt{e^2-1}}{e-1}>\frac{e^{-1}}{\sqrt{2}}.$$