# An elementary proof of the equivalence of measure theoretic and density expected values

Let $$(\Omega,\mathcal{F},P)$$ be a probability space, let $$X\colon \Omega\to\mathbb{R}$$ be real-valued and measurable. Suppose there exists $$f\colon \mathbb{R}\to [0,\infty]$$ such that $$P(X\in A)=\int_A f(x)\mathrm{d}x$$ for each $$A\in\mathrm{Borel}(\mathbb{R})$$.

I wish to show that $$\mathbb{E}[X]=\int X\mathrm{d}P=\int_{\mathbb{R}}xf(x)\mathrm{d}x$$, using the most elementary techniques possible (e.g. using indicator and simple functions, followed by integral limit theorems, as opposed to Radon-Nikodym, pushforward measures, etc.) Other answers I have found seem to use technical reasoning I am unfamiliar with.

One result I have proven is the following: Define $$\mu\colon\mathrm{Borel}(\mathbb{R})\to[0,1]$$ by $$\mu(A)=P(X\in A)$$. Then for any measurable function $$g\colon\mathbb{R}\to[0,\infty]$$, $$\mathbb{E}[g(X)]=\int_{\mathbb{R}}g \mathrm{d}\mu.$$

If we set $$g=\mathrm{id}$$ in the previous claim, then we obtain $$\mathbb{E}[X]=\int_{\mathbb{R}}\mathrm{d}\mu$$. Can we use the fact that $$\mu(A)=\int_A f(x)\mathrm{d}x$$ for each $$A\in\mathrm{Borel}(\mathbb{R})$$, to write $$\int_{\mathbb{R}} \mathrm{d}\mu=\int_{\mathbb{R}}xf(x)\mathrm{d}x$$.

• Your formula for the expectation is wrong. If $X$ has a density $f$, then $EX=\int xf(x)\, dx=\int x dF(x)$. Commented May 5, 2019 at 22:37

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In this answer (to my own question), I show how the elementary definition of an expectation of a continuous random variable is consistent with the abstract measure theoretic formulation, and also mention the appearance of the Radon-Nikodym derivative. I heavily use notation from David William's Probability with Martingales [PW], and also restate his Lemma from Section 6.12 along with results from Section 5.14 in [PW] with detailed proof.

## Definitions

• Let $$(\Omega,\F,\P)$$ be a probability space, where $$\Omega$$ is a non-empty set, $$\F$$ is a $$\sigma$$-algebra on $$\Omega$$, and $$\P\colon\F\to \R$$ is a probability measure.
• Let $$\B$$ be the Borel $$\sigma$$-algebra on $$\R$$.
• Let $$m\F$$ be the set of random variables, i.e., $$m\F=\{X\colon\Omega\to\R:X\text{ is }\F/\B\text{ measurable}\}.$$
• Let $$m\B$$ be the set of real-valued measurable functions on $$\R$$, i.e. $$m\B=\{h\colon\R\to\R:h\text{ is }\B/\B\text{ measurable}\}.$$
• Let $$\Lambda_X$$ be the law of $$X$$, which is the probability measure $$\Lambda_X\colon\B\to\R$$ such that $$\Lambda_X(B)=\P(X\in B)$$ for all $$B\in\B$$.
• Use $$\E$$ to denote integration of elements of $$m\F$$ with respect to the measure $$\P$$, e.g. $$\E[X]=\int_{\Omega} X \d \P$$.
• Let $$\mathbf{L}_X$$ be the integration operator (analagous to $$\E$$) of elements of $$m\B$$ with respect to $$\Lambda_X$$. That is, $$\mathbf{L}_X(h)=\int_{\R} h\d\Lambda_X$$ for each $$h\in m\B$$, where integration is with respect to $$\Lambda_X$$ instead of $$\P$$.
• Set $$\L^1(\Omega,\F,\P) = \{X \in m\F:\E|X|<\infty\},$$ the set of integrable random variables with respect to $$\P$$.
• Let $$\L^1(\R,\B,\Lambda_{X})= \{h\in m\B: \int_{\R}|h|\d\Lambda_X<\infty \}$$, the set of integrable functions on $$\R$$ with respect to $$\Lambda_X$$.
• Let $$\mathrm{Leb}$$ denote the Lebesgue measure, and we write $$\int_{\R} h \d\mathrm{Leb}$$ or $$\int_{\R}h(x)\d x$$ to denote the Lebesgue integral of $$h\in m\B$$.

Note that we are using two different notions of integral for $$\E$$ and $$\Lambda_X$$. That is, just as we define integration of $$\F/\B$$ measurable functions $$X\colon\Omega\to\R$$ with respect to the probability measure $$\P$$, we may define integration of $$\B/\B$$ measurable functions $$h\colon\R\to\R$$ with respect to the probability measure $$\Lambda_X$$. Instead of $$\int_{\Omega} X\d \P$$, we write $$\int_{\R} h \d \Lambda_X$$. Observe that $$\mathbf{L}_X(h)$$ relates to $$X$$ precisely through the measure $$\Lambda_X$$ we are integrating with respect to, while $$\P$$ has no connection a priori to $$X$$ in $$\E[h(X)]$$.

Our goal is to establish that the $$\mathbf{L}_X$$ and $$\E$$ operators agree (Proposition 1), i.e., $$\int_{\R} h\d\Lambda_X = \int_{\Omega}h(X)\d\P$$, and in the special case that the random variable admits a density, $$\mathbf{L}_X$$ has the standard elementary form $$\int_{\R}xf(x)\d x$$ (Proposition 3).

## Equivalence of Integration with respect to Law $$\Lambda_X$$ and with respect to $$\P$$

We state and prove the Lemma from Section 6.2 of [PW]. It states that a $$\B/\B$$-measurable function $$h$$ is integrable with respect to $$\Lambda_X$$ iff $$h(X)$$ is integrable with respect to $$\P$$, in which case the integrals agree. More compactly,

Proposition 1: Suppose $$h\in m\B$$. Then $$h(X)\in\L^1(\Omega,\F,\P)$$ iff $$h\in\L^1(\R,\B,\Lambda_X)$$ and $$\E h(X)= \mathbf{L}_X(h)$$, i.e. $$\int_{\R} h\d\Lambda_X = \int_{\Omega}h(X)\d\P$$.

Proof. First, suppose $$h=1_B$$ for $$B\in\B$$. Then $$\E h(X)=\E[1_{\{X\in B\}}] =P(X\in B),$$ and $$\mathbf{L}_X(h) =\int_{\R}1_B \d\Lambda_X = \int_B \d\Lambda_X = \Lambda_X(B) :=P(X\in B).$$

Linearity of abstract integration will demonstrate that equality holds for each simple function.

Next, let $$h$$ be a non-negative function in $$m\B$$. Then there is a sequence $$(h_n)$$ of non-negative simple functions in $$m\B$$ such that $$h_n\uparrow h$$. By the Monotone Convergence Theorem applied to each distinct measure space, $$\E[h(X)]=\lim_n \E[h_n(X)] = \lim_n \mathbf{L}_X(h_n)=\mathbf{L}_X(h),$$ since $$\E[h_n(X)]=\mathbf{L}_X(h_n)$$ for each $$n$$.

Most generally, if $$h\in m\B$$, write $$h=h^{+}-h^{-}$$ where $$h^{+}=\max(0,h)$$ and $$h^{-}=\max(-h,0)$$ and apply linearity.

## When a PDF Exists

Next, let's obtain a familiar expression for $$\mathbf{L}_X(h)$$ when $$X$$ admits a PDF $$f_X$$, i.e., $$\Lambda_X(B)=P(X\in B)=\int_B f_X\d \mathrm{Leb} = \int_B f_X(x)\d x.$$ In the notation of [PW], this means $$\Lambda_X$$ has measure $$f_X\mathrm{Leb}$$, $$\Lambda_X$$ has density $$f_X$$ relative to $$\mathrm{Leb}$$, and we write $$\frac{\d \Lambda_X}{\d \mathrm{Leb}}=f_X$$. The PDF $$f_X$$ is called the Radon-Nikodym derivative of $$\Lambda_X$$ relative to $$\mathrm{Leb}$$ on $$(\R,\B)$$. Our precise formulation is below, but more suggestively, we demonstrate that $$\int_{\R} h(x)f_X(x)\d x=\int_{\R} h f_X\d\mathrm{Leb} = \int_{\R} h \frac{\d \Lambda_X}{\d \mathrm{Leb}}\d\mathrm{Leb} = \int_{\R} h\d\Lambda_X.$$

Proposition 2: Suppose $$h\in m\B$$. Then $$h(X)\in\L^1(\Omega,\F,\P)$$ iff $$h\in\L^1(\R,\B,\Lambda_X)$$. Assume $$h\in \L^1(\R,\B,\Lambda_X)$$ where $$\Lambda_X$$ has density $$f_X$$ relative to $$\mathrm{Leb}$$. Then $$\int_{\R} h\d\Lambda_X =\int_{\R}hf_X\d\mathrm{Leb}.$$

Proof. First, suppose $$h=1_C$$ where $$C\in\B$$. Then $$\int_{\R}h\d\Lambda_X = \int_C \d\Lambda_X =\Lambda_X(C),$$ and $$\int_{\R}hf_X\d\mathrm{Leb} = \int_C f_X\d\mathrm{Leb} =P(X\in C)=\Lambda_X(C).$$ Our assertion immediately extends to simple functions using linearity properties of both integrals.

Suppose $$h\in\L^1(\R,\B,\Lambda_X)$$ with $$h\ge0$$. Then there is an increasing sequence $$(h_n)$$ of non-negative simple functions such that $$h_n\uparrow h$$. By the Monotone Convergence Theorem applied to each distinct measure space, \begin{align*} \int_{\R}h\d\Lambda_X =\lim_n\int_{\R}h_n\d\Lambda_X=\lim_n\int_{\R} h_nf_X\d\mathrm{Leb}= \int_{\R} hf_X\d\mathrm{Leb}, \end{align*} yielding the claim. For general $$h\in\L^1(\R,\B,\Lambda_X)$$, writing $$h=h^{+}-h^{-1}$$ finishes the proof.

The following final result combines Proposition 2 with Proposition 1 to yield $$\int_{\Omega} h(X)\d \P=\int_{\R} h(x)f_X(x)\d x,$$ where the latter integral is the Lebesgue integral.

Proposition 3: Suppose $$h\in m\B$$. Then $$h(X)\in\L^1(\Omega,\F,\P)$$ iff $$h\in\L^1(\R,\B,\Lambda_X)$$. Moreover, if $$\Lambda_X$$ has density $$f_X$$ relative to $$\mathrm{Leb}$$, then $$\E[X]=\int_{\Omega} h(X)\d \P =\int_{\R}hf_X\d\mathrm{Leb}.$$

## References

Williams, D. (1991). Probability with Martingales. Cambridge: Cambridge University Press. doi:10.1017/CBO9780511813658