Problem 4

Exercise 5.4 (Absolutely continuous measures)   Let $(X,\Sigma,\mu)$ be a finite measure space and suppose $f,g>0$. Define the measures

$$\nu(E) = \int_E fd\mu \quad \eta(E) = \int_E gd\mu$$ (5.14)

Is $\nu \ll \eta$? Is $\eta \ll \nu$?

Proof. Compute $R=\operatorname{ess}\sup g(x)/f(x)$. This is finite because $X$ is a finite measure space and $f$ and $g$ are strictly positive.

If $\nu(E)=0$, we can show $\eta(E)=0$. By the definition of the Lebesgue integral, select

$$f \leq \sum_{k=1}^n c_k\mathbf{1}_{E_k}$$ (5.15)

satisfying

$$\int_E f \leq \int \sum_{k=1}^n c_k\mathbf{1}_{E_k} < \epsilon/R$$ (5.16)

By the selection of $R$, it can be readily seen that

$$\frac{g(x)}{f(x)} \leq R \implies g(x) \leq Rf(x)$$ (5.17)

Apply the monotonicity of the integral to find

$$\int_E g \leq R\int_E f < R\epsilon/R = \epsilon$$ (5.18)

Since $\epsilon>0$ is arbitrary, this implies $\eta(E)=0$, so that $\eta \ll \nu$. A similar argument can be made with $R' = 1/R$ to show that $\nu \ll \eta$. $\qedsymbol$