Dear Mathematics community,
Def 4.1 ($ M_1 $) Obviously it lives up to $ M_1 $ as any measure of the emptyset is zero, $ \rho(\emptyset)=a\mu(\emptyset)+b\nu(\emptyset)=a\cdot 0+b\cdot 0=0 $ ($ M_2 $) Take <$ \{A_j\}_{j\in\mathbb{N}}\subset \mathcal{F} $ be disjoint sets. Then using $ \sigma $-additivity we can see: \begin{equation*} \rho(\cup_{j\in\mathbb{N}}A_j)=a\mu(\cup_{j\in\mathbb{N}}A_j)+b\nu(\cup_{j\in\mathbb{N}}A_j)=a\sum_{j\in\mathbb{N}}\mu(A_j)+b\sum_{j\in\mathbb{N}}\mu(A_j) \end{equation*} \begin{equation} =\sum_{j\in\mathbb{N}}(a\mu(A_j)+b\nu(A_j)=\sum_{j\in\mathbb{N}}\rho(A_j) \end{equation} \end{proof} I am doing some problems in measure theory and ran into problems with two problems. I sence that the problem is a lack of understanding in measures but maybe a clarification can help. \begin{equation}\tag{1} \rho(A):=a\mu(A)+b\nu(A)\quad \text{for }A\in\mathcal{F} \end{equation} Let $ \rho=a\mu+b\nu $ be a measure defined by (1). Prove that if a function $ f : X \rightarrow \mathbb{R} $ is both $ \mu $-integrable and $ \nu $-integrable, then f is also $ (a\mu + b\nu) $-integrable and we have \begin{equation} \int fd(a\mu+b\nu)=a\int fd\mu+b\int fd\nu \end{equation} My attempt: Assuming that $ f\geq0 $, we can then find a monotonically increase (or decreasing) sequence of non-negative, measurable functions $ \{f_n\}_{n=1}^\infty $, such that $ \lim\limits_{n\rightarrow\infty}f_n=f $. Using the Monotone convergence theorem twice we can get the result that we ar looking for. \begin{equation} \int fd\rho=\int \lim\limits_{n\rightarrow\infty}f_n d\rho= \lim\limits_{n\rightarrow\infty}\int f_n d\rho= \lim\limits_{n\rightarrow\infty} \int f_nd(a\mu+b\nu)= \end{equation} \begin{equation} \lim\limits_{n\rightarrow\infty} a\int f_nd\mu+\lim\limits_{n\rightarrow\infty} b\int f_nd\nu=a\int fd\mu+b\int fd\nu \end{equation} I am unsure if it is permitted to do what I have from line 1 to 2? Second problem: \begin{equation}\tag{2} f(x)=\begin{cases} |x|^{-1/2}=\frac{1}{\sqrt{|x|}},& if x\neq 0\\ \infty, & if x=0 \end{cases} \end{equation} Let $ f:\mathbb{R}\rightarrow\bar{\mathbb{R}}$ be the function given by (2). Consider thge measure $ \rho_t=\lambda^1+\delta_t $ where ,$ \lambda^1 $ is the one-dimensional Lebesgue measure and $ \delta_t $ is tge Dirac measure accumulated in a point $ t\in \mathbb{R} $. Calculate \begin{equation} \int_{[0,1]}fd\rho_t \end{equation} For which $ t $ the function $ f $ is $ \rho_t $-integrable on the interval $ [0,1] $? Here i guess I will be able to use the first problem to split the integra into two: \begin{equation} \int_{[0,1]}fd\rho_t =\int_{[0,1]}fd\lambda^1+\int_{[0,1]}fd\delta_t \end{equation} But from here i have no idea.