# Measure Defined by Lebesgue Integral Of Another Measure

Let $$(\Omega, \mathcal{A}, \mu)$$ be a measure space and $$f: \Omega \to \mathbb{R}$$ a measure nonnegative function that is summable against $$\mu$$.

It is known that $$\nu(A) = \int_{A} f \mu$$ is also a measure on $$\mathcal{A}$$. I have to prove that $$\int_{\Omega} g \nu = \int_{\Omega} gf \mu$$ for all functions $$g: \Omega \to \mathbb{R}$$ that are summable against $$\nu$$.

I don't quite understand the notation, though. Specifically, on the proof statement, what does $$gf \mu$$ mean? Am I taking the composition of $$g$$ and $$f$$? How does one even evaluate $$\int_{\Omega} g \nu$$?

Thanks in advance for any help.

• The more common notation is $\nu(A)=\int_Af\,d\mu$; the integral of $f$ with respect to the measure $\mu$. You're then supposed to show $\int_{\Omega}g\,d\mu=\int_{\Omega}g\cdot f\,d\mu$; the RHS has a product of $g$ and $f$. In words: show Lebesgue integral of $g$ with respect to $\nu$ equals Lebesgue integral of $gf$ (a product) with respect to $\mu$. These are all Lebesgue integrals with respect to the measures involved, so the definition goes like non-negative simple functions $\to$ non-negative measurable functions $\to$ real integrable $\to$ complex-valued integrable functions. Mar 24, 2022 at 1:37
• So to start off with simple functions, if I just consider indicator function, $g = a * X_A,$, $$\int g \nu = a \cdot \nu (A) = a \cdot \int_{A} f \mu = a \cdot \int X_{A} f \mu = \int a X_{A} f \mu = \int gf \mu.$$ Does that work? Mar 24, 2022 at 1:45
• Yes that's exactly right. Now use linearity to get all simple functions, and monotone convergence to get all non-negative measurable functions. Mar 24, 2022 at 1:57
• @peek-a-boo does "all functions summable against $\nu$" mean I have to go further than non-negative measurable functions? Do I need Dominated Convergence Theorem like how Mahdi said? Or is this much enough? Mar 24, 2022 at 1:59
• if you know the equality for all non-negative functions then for integrable functions the statement becomes clear by considering positive and negative parts. Mar 24, 2022 at 2:31

Regarding your questions on the notation $$\nu(A)=\int f d\mu$$ and $$\int g d \nu=\int fg d \mu$$, and in particular $$fg$$ means multiplication of $$f$$ and $$g$$.
The question you are asking can be answered using Radon Nikodym theorem. However, you can also directly approach this question by assuming first that $$g$$ is a simple function. Then using the fact that you can approximate any measurable function with simple functions and dominated convergence theorem you can generalize it to any function.