Measure with density

Remarks

The map $f\mu$ is a measure: $(f\mu)(\emptyset) = 0$, and countable additivity follows from the monotone convergence theorem applied to $f \mathbf{1}_{A_1 \cup \dots \cup A_n}$. It is absolutely continuous with respect to $\mu$, since $\mu(A) = 0$ implies $\int_A f \, d\mu = 0$.

Example: on $\R$ with Lebesgue measure and $f = \mathbf{1}_{[0,1]}$, the measure $f\mu$ is Lebesgue measure restricted to $[0, 1]$. Check: $(f\mu)(A) = \int \mathbf{1}_{A \cap [0,1]} \, dx$, the length of $A \cap [0, 1]$.

Non-example: $f(x) = x$ on $\R$ takes negative values, so it is not an admissible density. The formula $A \mapsto \int_A x \, dx$ is not even defined for $A = \R$.

For measurable $g \ge 0$, $\int g \, d(f\mu) = \int g f \, d\mu$. Notation: $f\mu$, $f \cdot \mu$, or $d\nu = f \, d\mu$.