# Relationship between induced measure and measure corresponding to a density function.

I am reading these lecture notes, section 2.3 (pg 4), and I've become very confused about relationship between

1. The induced measure $$\mu_X$$ -- a random variable $$X: \Omega \rightarrow S$$ with the original measure $$\mu$$ induces $$\mu_X(B) = \mu(X^{-1}(B))$$.

2. The measure $$\nu(A) = \int_A f\ d\mu$$ corresponding to a density function $$f:\Omega \rightarrow \mathbb{R}^{0+}$$.

The notes compare the definition of density function I'm familiar with: $$Pr(X \leq a) = \int_{-\infty}^a f(x)\ dx$$ with the measure-theoretic equivalent: $$\mu_X(B) = \int_B f\ d\lambda$$ where $$\lambda$$ is the Lebesgue measure.

I'm trying to reconcile the original definition of $$\mu_X$$ with the new one, and I can't see why it should be the case that $$\mu_X(B) = \int_B f\ d\lambda = \int_{X^{-1}(B)}1\ d\mu$$ where the RHS is just another way of writing the original $$\mu(X^{-1}(B))$$.

I'm also confused because $$\nu$$ and $$\mu_X$$ are written so similarly you'd suspect they're the same thing, but $$\mu_X$$ is a measure on $$(S, \mathcal{A})$$ whereas $$\nu$$ is a measure on $$(\Omega, \mathcal{F})$$, so this is clearly not possible. But then I'm not sure what was the point of $$\nu$$.

I think this is somehow related to one of the theorems stated in the text: $$\int g\ d\nu = \int f g\ d \mu$$ (when $$f, \nu, \mu$$ are related as described above), but if I use this by plugging in $$\lambda \rightarrow \mu, I_B \rightarrow g$$ ($$I_B$$ being the indicator function for set $$B$$)) I end up with $$\int I_B d\nu$$ where $$\nu(A) = \int_A1\ d\lambda$$, which still doesn't get me anything I want.

What am I missing?

For any random variable $$X$$ we always have a measure $$\mu_X$$ on the Borel sigma algebra of the real line defined by $$\mu_X(B)=P(X^{-1}(B))$$. In general there is no density function of $$f$$. We say that $$X$$ has a density if there exist a non-negative measurable function $$f$$ such that $$P(X^{-1}(B)=\int_B f(x)dx$$ for all Borel sets $$B$$. In 2) $$\mu$$ is not $$\mu_X$$ but it is the Lebesgue measure. $$\nu$$ is same as $$\mu_X$$ and we have $$\mu_X(B)=\int_B f(x)dx$$ for all Borel sets $$B$$.
• i'm a bit confused by $\int_B f(x) dx$ here -- is this the same as $\int_B f d\lambda$ ($\lambda$ being the Lebesgue measure), or $\int_{-\infty}^a f(x) dx$, or something else altogether? Commented Jun 19, 2020 at 15:06
• @shimao $dx$ is same as $d \lambda (x)$ where $\lambda$ is Lebesgue measure. Commented Jun 19, 2020 at 23:11