Is integral linear with respect to measures? In particular, scalar measures? I got an exercise that asked to show the following:
Statement: Given two scalar measures $\lambda_1, \lambda_2$ (either real or complex) and a scalar $k$. Show that if $f \in L^1(\lambda_1)\cap L^2 (\lambda_2)$, then $f \in L^1(k\lambda_1 + \lambda2)$ and
$$\int_A fd(k \lambda_1 + \lambda_2) = k \int_A fd\lambda_1 + \int_A fd\lambda_2$$
Note: I've already shown this result for non-negative, finite measures and non-negative $k \in \mathbb{R}$. However, using the definition of the integral for scalar measures, I don't see how to proceed. That is, I have that
$$\int_A fd(k \lambda_1 + \lambda_2) = \int_A fd(k \lambda_1 + \lambda_2)^+ - \int_A fd(k \lambda_1 + \lambda_2)^-$$
where $(k \lambda_1 + \lambda_2)^+$ and $(k \lambda_1 + \lambda_2)^-$ is the Jordan-Hahn decomposition of $k \lambda_1 + \lambda_2$. Nonetheless, I don't think that the Jordan decomposition is linear. I also tried using the Radon-Nikodym derivative, but I also get stuck.
 A: Okay, let's just clear out that confusion.
First claim: If $\mu$ and $\nu$ are two signed measures, then for any $f\in L^1(\mu)\cap L^1(\nu),$ we have $f \in L^1(\mu+\nu)$ and $\int f\textrm{d}(\mu+\nu)=\int f\textrm{d}\mu+\int f\textrm{d}\nu.$ You are correct that the Jordan decompositions need not be compatible. However, assume that $\mu=\mu^+-\mu^-$ and $\nu=\nu-\nu^{-}$ are the Jordan decompositions of $\mu$ and $\nu$ respectively. Then, for any indicator function $B$, we have, by definition that
\begin{align}
\int1_B\textrm{d}(\mu+\nu)^+-\int 1_B \textrm{d}(\mu+\nu)^-&=
\int 1_B \textrm{d}(\mu+\nu)=(\mu+\nu)(B)= \mu^{+}(B)-\mu^{-}(B)+\nu(B)-\nu^{-}(B) \\&=\int 1_B\textrm{d}\mu^+-\int 1_B\textrm{d}\mu^-+\int 1_B\textrm{d}\nu^+-\int 1_B\textrm{d}\nu^-
\end{align}
By linearity of the left-hand side and right-hand side in $f$, we get the identity for all simple functions - so far we haven't even used that all of these measures were positive. Of course, for general functions, we need the definition $\int 1_B \textrm{d}(\mu+\nu)=\int 1_B \textrm{d}(\mu+\nu)^+-\int 1_B \textrm{d}(\mu+\nu)^-$.
Now, if $f$ is any positive, measurable function, such that $f\in L^1(\mu)\cap L^1(\nu)$, pick a sequence of simple function $f_n,$ which increases to $f$ pointwise and let $\Omega^+$ and \Omega^{-} be sets such that $(\mu+\nu)^+(\Omega^-)=0$ and vice versa. Then, we can apply the monotone convergence theorem to each of these six positive measures, to get
\begin{align}
\int 1_{\Omega^+}f\textrm{d}\mu+\int 1_{\Omega^+}f\textrm{d}\nu&=\int 1_{\Omega^+} f\textrm{d}\mu^+-\int 1_{\Omega^+}f\textrm{d}\mu^-+\int 1_{\Omega^+}f\textrm{d}\nu^+-\int 1_{\Omega^+}f\textrm{d}\nu^- \\
&=\lim_{n\to\infty} \left(\int 1_{\Omega^+}f_n\textrm{d}\mu^+-\int 1_{\Omega^+}f_n\textrm{d}\mu^-+\int 1_{\Omega^+}f_n\textrm{d}\nu^+-\int 1_{\Omega^+}f_n\textrm{d}\nu^-\right) \\
&=\lim_{n\to\infty}\left(\int 1_{\Omega^+}f_n\textrm{d}(\mu+\nu)^+-\int 1_{\Omega^+}f_n\textrm{d}(\mu+\nu)^-\right)\\
 &=\int 1_{\Omega^+}f\textrm{d}(\mu+\nu)^+-\int 1_{\Omega^+}f\textrm{d}(\mu+\nu)^-\\
&=\int 1_{\Omega^+}f\textrm{d}(\mu+\nu),
\end{align}
where we avoid a potential $\infty-\infty$ scenario since $\int 1_{\Omega^+}f\textrm{d}(\mu+\nu)^-=0$. Now, we can apply a similar argument to $(1_{\Omega^-})f$ in order to apply linearity to get that $\int f\textrm{d}(\mu+\nu)=\int f\textrm{d}\mu+\int f\textrm{d} \nu$
Now, we get the general result by splitting up $f=f^+-f^-$ for a general $L^1$ function.
Finally, the case where we apply multiply by a scalar is completely similar, and of course, the same is true for complex measures.
