Prove that $X$ has density $f$ if and only if $\text E[h(X)]=\int h(x)f(x)\;{\rm d}x$ for all Borel measurable bounded $h:\mathbb R\to\mathbb R$

Question

Let $X$ be a real-valued random. How can we prove that $X$ has density $f$ if and only if $$\operatorname E[h(X)]=\int h(x)f(x)\;{\rm d}x$$ for all Borel measurable bounded $h:\mathbb R\to\mathbb R$?

The direction "$\Rightarrow$" is clear by the law of the unconscious statistician (even when they don't state that the $h$ need to be bounded, $h(X)$ won't be integrable otherwise), but how can we show the other direction?

Answer 1

Proof of the law of the unconscious statistician. $∫_A f(x) \ dx = ∫\Bbb 1_A f(x) \ dx = \Bbb P(X ∈ A) = \Bbb E \Bbb 1_A$. Linearity tells us that the result then holds for all simple functions $h = ∑a_i \Bbb 1_{A_i}$, which proves the result by the density of simple functions.

From the first line of the above proof (i.e. take $h = \Bbb 1_A$ for any Borel $A$) we recover the direction that you are asking for (i.e. if $\Bbb E h(X) = ∫ h(x) f(x) \ dx$, then for every Borel set $A$, $\Bbb P(X∈ A) = ∫_A f dx$)

Answer 2

It's way more easy than I thought. Let's start with "$\Rightarrow$":

• Let $h:\mathbb R\to\mathbb R$ be bounded and Borel measurable $\Rightarrow$ $g(X)\in L^1(\operatorname P)$ with $$\operatorname E[g(X)]=\int g\;{\rm d}\operatorname P_X$$
• $X$ has density $f$ $\Rightarrow$$^1$ $\operatorname P_X=f\lambda^1$ and hence $$\int g\;{\rm d}\operatorname P_X=\int fg\;{\rm d}\lambda^1$$

Now, let's prove "$\Leftarrow$":

• By assumption (and with the argumentation of the proof of direction "$\Rightarrow$" in mind), $$\int g\;{\rm d}\operatorname P_X=\operatorname E[g(X)]=\int fg\;{\rm d}\lambda^1\stackrel{\text{def}}=\int g\;{\rm d}(f\lambda^1)$$ for all bounded and Borel measurable $g:\mathbb R\to\mathbb R$
• Choose $g=1_A$ for some$^2$ $A\in\mathcal B(\mathbb R)$ $\Rightarrow$ $$\operatorname P_X[A]=(f\lambda_1)(A)$$
• Since $A$ was arbitrary, we obtain $$\operatorname P_X=f\lambda_1$$

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$^1$ Let $\lambda^1$ denote the Lebesgue measure on $\mathbb R$ and $f\lambda^1$ denote the measure with density $f$ with respect to $\lambda^1$.

$^2$ Let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$.