## Measure that is absolutely continuous with respect to mu

Interesting Career Personality Test (Free): https://mathtuition88.com/free-career-quiz/

Let $(X,\mathcal{M},\mu)$ be a measure space, and let $f:X\to [0,\infty]$ be a measurable function. Define the map $\lambda:\mathcal{M}\to[0,\infty]$, $\lambda(E):=\int_X \chi_E f d\mu$, where $\chi_E$ denotes the characteristic function of $E$.

(a) Show that $\lambda$ is a measure and that it is absolutely continuous with respect to $\mu$.

(b) Show that for any measurable function $g:X\to[0,\infty]$, one has $\int_X g d\lambda=\int_X gf d\mu$ in $[0,\infty]$.

Proof: For part (a), we routinely check that $\lambda$ is indeed a measure. $\lambda(\emptyset)=\int_X \chi_\emptyset f d\mu=\int_X 0 d\mu=0$. Let $E_i$ be mutually disjoiint measurable sets. \begin{aligned} \lambda(\cup_{i=1}^\infty E_i)&=\int_X \chi_{\cup_{i=1}^\infty E_i} f d\mu\\ &=\int_X (\sum_{i=1}^\infty \chi_{E_i}) f d\mu\\ &=\sum_{i=1}^\infty \int_X \chi_{E_i} f d\mu\\ &=\sum_{i=1}^\infty \lambda (E_i) \end{aligned}

If $\mu (E)=0$, then $\chi_{E} f=0$ a.e., thus $\lambda (E)=0$. Therefore $\lambda\ll\mu$.

(b) We note that when $g$ is a characteristic function, i.e. $g=\chi_E$, \begin{aligned} \int_X g d\lambda&=\int_X \chi_E d\lambda\\ &=\lambda (E)\\ &=\int_X \chi_E f d\mu\\ &=\int_X gf d\mu \end{aligned}

Hence the equation holds. By linearity, we can see that the equation holds for all simple functions. Let $(\psi_n)$ be a sequence of simple functions such that $\psi_n\uparrow g$. Then by the Monotone Convergence Theorem, $\lim_{n\to\infty} \int \psi_n d\lambda=\int g d\lambda$.

Note that $\psi_n f\uparrow gf$, thus by MCT, $\lim_{n\to\infty}\int\psi_n f d\mu=\int g f d\mu$. Note that $\int \psi_n d\lambda=\int \psi_n f d\mu$. Hence, $\int g d\lambda=\int gf d\mu$, and we are done. 