I know that for $f \in L^1$, and $\mu(S) > 0$, we have $$\int_X f \, d\mu = \int_{X \setminus S} f \, d\mu + \int_{S} f \, d\mu.$$ Is the same true in $L^p$? Or do we now get an inequality? Namely, does
$$\left(\int_X f^p \, d\mu \right)^{\frac{1}{p}} = \left(\int_{X \setminus S} f^p \, d\mu \right)^{\frac{1}{p}}+\left(\int_S f^p \, d\mu \right)^{\frac{1}{p}}?$$
I'm trying to prove $L^p$ is separable by showing the collection$$ S:= \{\sum_{i=1}^nr\chi_{(a_i,b_i)}\}_{(a_i,b_i,r) \in \mathbb{Q}^3}$$ is dense in $L^p$, which leads me to the estimate, given $\chi_E$, $E$ measurable, and a member of $S$ with $r=1$, then $$\left(\int_X |\chi_E - \chi_{(a,b)}| \, d\mu \right)^{\frac{1}{p}} = \left(\int_{E \setminus (a,b)} (1) \right)^\frac{1}{p} + \left(\int_{(a,b) \setminus E} (1) \right)^\frac{1}{p} $$ since $|\chi_E - \chi_{(a,b)}|$ is $0$ except on $(a,b) \setminus E \cup E \setminus (a,b)$. Also, am I correct in saying that if I can get the difference of the coefficients to add up to less than $1$ and the difference in the measure of the sets for the simple function to add up to less than $\epsilon^p$ then I've proven density?
No it is not true. Generally $$ \left(\int_X f^p \, d\mu \right)^{\frac{1}{p}} = \left(\int_{X \setminus S} f^p \, d\mu+ \int_{S} f^p \, d\mu\right)^{\frac{1}{p}}\ne\left(\int_{X \setminus S} f^p \, d\mu \right)^{\frac{1}{p}}+\left(\int_S f^p \, d\mu \right)^{\frac{1}{p}} $$ By Minkowski inequality $$ \left(\int_X |\chi_E - \chi_{(a,b)}| \, d\mu \right)^{\frac{1}{p}} \leqslant \left(\int_{E \setminus (a,b)} (1) \right)^\frac{1}{p} + \left(\int_{(a,b) \setminus E} (1) \right)^\frac{1}{p} $$