Show that the integral against measures is additive in the measure Let $\nu_i, i = 1,2$, be measures on $(\Sigma, \mathcal{F})$ and $f$ be Borel,  show that  $\int f\, d(\nu_1+\nu_2) = \int f\,d\nu_1 + \int f\, d\nu_2$, i.e., if either side of the equality is well defined, then so is the other side, and the two sides are equal.
for me, I think I can use simple function to prove LHS to RHS, but how can I prove RHS to LHS.
\begin{align}
 \int  f  d(\nu_1+\nu_2)  &= \int f^{+} d(\nu_1+\nu_2) -\int f^{-} d(\nu_1+\nu_2)   \nonumber\\
 &=  \sum_{i = 1}^{k} a_i(\nu_1 + \nu_2)(A_i) - \sum_{i = 1}^{k}b_i(\nu_1 + \nu_2)(A_i) \nonumber \\
 &= \sum_{i = 1}^{k} ( a_i\nu_1 (A_i) + a_i \nu_2(A_i) ) - \sum_{i = 1}^{k} (b_i \nu_1 (A_i) + b_i \nu_2(A_i) ) \nonumber \\
 &= \sum_{i = 1}^{k}  a_i\nu_1 (A_i) - \sum_{i = 1}^{k} b_i \nu_1 (A_i) + \sum_{i = 1}^{k}  a_i\nu_2 (A_i) - \sum_{i = 1}^{k} b_2 \nu_1 (A_i) \nonumber \\
 &= \int f\,d\nu_1 + \int f\, d\nu_2 \nonumber
\end{align}
 A: With these kinds of equalities, it often suffices to prove it for measurable $f \colon \Sigma \to [0, \infty]$ because of the way the integral is defined. For this proof, this strategy happens to work nicely.
We first show that $\int f\,d(\nu_1 + \nu_2) = \int f\,d\nu_1 + \int f\,d\nu_2$ for every measurable $f \colon \Sigma \to [0, \infty]$. This is easy enough for simple $f$. By the monotone convergence theorem (take simple $\phi_n \nearrow f$), this holds for all measurable $f \colon \Sigma \to [0, \infty]$.
It follows that $L^1(\Sigma, \nu_1 + \nu_2) = L^1(\Sigma, \nu_1) \cap L^1(\Sigma, \nu_1)$. By taking positive and negative parts of $f$ and using the linearity of the integral on $L^1(\Sigma, \nu_j)$, it follows the equality also holds for all real valued $f \in L^1(\Sigma, \nu_1 + \nu_2)$. By taking real and imaginary parts and using linearity of the integral on $L^1(\Sigma, \nu_j)$, it follows that the equality holds for all $f \in L^1(\Sigma, \nu_1 + \nu_2)$.
Thus the equality holds for all measurable $f \colon \Sigma \to [0, \infty]$ and for all $f \in L^1(\Sigma, \nu_1 + \nu_2)$.
