# How to show for positive Borel functions $g, \int_{a}^{b} g(x)dF(x) = \int_{a}^{b} g(x)f(x)dx$

$$\textbf{Problem:}$$ Let $$\mu$$ be an absolutely continous probability distribution with distribution function $$F(x)$$ and density $$f(x)$$. For any Borel set $$\Lambda$$ and $$a, let $$\nu\left(\Lambda \cap[a,b] \right) = \int_{a}^{b} I_\Lambda (x) f(x) dx$$. Show that this defines a probability measure $$\nu$$ on the Borel sets of $$\mathbb{R}$$, and that $$\nu = \mu$$ . Conclude that for positive Borel functions $$g, \int_{a}^{b} g(x)dF(x) = \int_{a}^{b} g(x)f(x)dx.$$

I am aware of the following theorem

$$\textbf{Theorem:}$$ Let $$X$$ be a random variable with distribution function $$F$$, and let $$g$$ be a positive Borel function. Then

$$\mathbb{E}\{g(X)\} = \int_{-\infty}^{\infty} g(x)dF(x).$$

In particular $$\mathbb{E}\{X\} = \int_{-\infty}^{\infty} x dF(x).$$

From a different problem I have seen that it might be useful to approach the problem by letting $$\mathcal{G}$$ be the class of sets on which $$\nu = \mu$$ since then $$(a,b] \in \mathcal{G}$$, but I could use some help and direction in how to put this together properly. I for some reason can't seem to make it work and my head is now hurting.

• Are you sure you have the right hand side of the equality correct? Commented Sep 26, 2022 at 1:04
• @Glen_b No, an extra d had gotten in there somehow. I have fixed it now. Thanks Commented Sep 26, 2022 at 2:46
• You need to define a bunch of terms here.
– Ben
Commented Sep 26, 2022 at 3:02
• what is $\mu$ ? And are the $a$ and $b$ in your definition of $\nu$ allowed to be $\pm \infty$ ? Commented Sep 26, 2022 at 11:56
• This is an exercise in chasing mathematical definitions. Thus, the solution you need will depend on which definitions you need to apply. In particular, please let us know what kind of integrals these are (Stieltjes? Lebesgue-Stieltjes? Henstock-Kurzweil? Other?).
– whuber
Commented Sep 26, 2022 at 13:26

Albeit the concerns re the ambiguities as shown in the comments deserve clarification, these sort of problems are standard measure theoretic exercises. So, I am leaving below a general brief discussion of the problem in hand following which OP can write the proofs formally.

Proposition $$1.1.$$ Let $$(\Omega, \mathfrak A, \mu)$$ be a measure space and let $$g$$ be a non-negative integrable function over $$\Omega.$$ Then the set function $$\mathfrak A\mapsto \mathbb R$$ $$\nu(B):= \int_B g~\mathrm d\mu,~~\forall B\in\mathfrak A\tag 1$$ is a measure on $$(\Omega, \mathfrak A).$$

It is obvious $$\nu(\emptyset) =0.$$ For countable additivity, consider a sequence of disjoint sets $$\langle B_i\rangle_{i\in\mathbb N},~B_i\in\mathfrak A.$$ Using $$\chi_{\cup B_i}= \sum_i \chi_{B_i},$$

\begin{align}\nu\left(\bigcup B_i\right) &= \int g\chi_{\cup B_i}~\mathrm d\mu\\&= \sum_i\int g\chi_{B_i}~\mathrm d\mu\\&= \sum_i\nu(B_i).\tag 2\end{align}

$$\square$$

Proposition $$1.2.$$ If $$f$$ is an extended real-valued measurable function on $$\Omega,$$ then $$\int_\Omega f~\mathrm d\nu = \int_\Omega fg~\mathrm d\mu.\tag 3$$

It suffices to show for a simple function $$\varphi=\sum_{i=1}^n a_i\chi_{B_i}:$$

\begin{align}\int \varphi~\mathrm d\nu &= \sum_{i=1}^n a_i\nu(B_i)\\ &= \sum_{i=1}^n a_i\int g\chi_{B_i}~\mathrm d\mu\\ &=\int \sum_{i=1}^n a_i \chi_{B_i}g~\mathrm d\mu\\&= \int \varphi g ~\mathrm d\mu.\tag 4 \end{align}

If $$f$$ is non-negative, then by Simple Approximation Lemma (cf. $$\rm [I],$$ section $$18.1,$$ p. $$363$$) there does exist an increasing sequence of simple functions $$\langle \varphi_i\rangle_{i\in\mathbb N}$$ converging to $$f;$$ then $$\varphi_i g \to f g.$$ The rest follows by Monotone Convergence Theorem (cf. $$\rm [I],$$ section $$18.2,$$ p. $$370$$).

$$\square$$

Proposition $$2.$$ Let $$(\Omega_1,\mathfrak A, \mu)$$ be a measure space and let $$(\Omega_2,\mathfrak B)$$ be a measure space. Let $$f$$ be $$\mathfrak A/\mathfrak B$$ measurable. If an extended real-valued $$g$$ is measurable on $$\Omega_2,$$ then $$g ~~\mu f^{-1}-\text{integrable}\iff g\circ f~~\mu-\text{integrable}.\tag{5.i}$$ Also, then $$\int_{\Omega_2} g~\mathrm d\left(\mu f^{-1}\right) =\int_{\Omega_1} (g\circ f )~\mathrm d\mu.\tag{5.ii}$$

Consider $$g$$ to be a characteristic function of $$B\in\mathfrak B.$$ Then $$g\circ f$$ would be the characteristic function of $$f^{-1}(B).$$ $$\rm (5.ii)$$ then holds as both are $$\mu\left(f^{-1}(B)\right).$$ It holds for non-negative measurable $$g$$ - resort again to Simple Approximation Lemma and Monotone Convergence Theorem. For the general $$g,$$ it holds true as $$g= g^+-g^-$$ (from $$[\rm II],$$ section $$2.6,$$ pp. $$75-76$$).

$$\square$$

• If $$\nu(\Omega) =1,$$ then $$\nu$$ is probability measure.

• $$g$$ is a density function which is essentially unique provided $$\mu$$ is $$\sigma$$-finite; also check Radon-Nikodym Theorem (cf. $$\rm[I],$$ section $$18.4,$$ p. $$382$$).

• Let $$(\Omega_1,\mathfrak A, \mathbb P)$$ be a probability space and let X is $$\mathfrak A/\mathfrak B$$ measurable. Then $$\mathbb P\circ X^{-1},$$ the distribution is a measure on $$(\Omega_2, \mathfrak B).$$ Let $$\mu$$ be a $$\sigma$$-finite measure on $$(\Omega_2, \mathfrak B).$$ If the distribution of $$X$$ is absolutely continuous w.r.t $$\mu,$$ the associated density being $$f,$$ and if $$g$$ is an extended real-valued measurable function on $$\Omega_2,$$ then, if it is integrable, using the above results,

\begin{align}\int_{\Omega_1} g\circ X ~\mathrm d\mathbb P &= \int_{\Omega_2} g~\mathrm d \left(\mathbb P\circ X^{-1}\right) \\ &= \int_{\Omega_2} g f~\mathrm d \mu.\tag 6 \end{align}

## References:

$$[\rm I]$$ Real Analysis, H. L. Royden, P. M. Fitzpatrick, Pearson Education, $$2010.$$

$$\rm [II]$$ Measure Theory, Donald L. Cohn, Springer Science$$+$$Business Media, $$2013.$$