Measure that is absolutely continuous with respect to mu

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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.


About mathtuition88
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