I am trying to understand the proof for the following statement:
Let $X$ be a random variable with $\underbrace{\text{well-defined expectation}}_{\mathbb{E} X = \int_\Omega X(\omega) \,d\mathbb{P}(\omega)}$ $(\Omega, \Sigma, \mathbb{P})$. Then we have $$ \mathbb{E} X = \int_\mathbb{R} t \,d\mu_X(t), $$ where $\mu_X(B) = \mathbb{P}\{ X \in B \}$ is the induced probability of $X$.
Here is the proof:
If $X$ is any non-negative variable, then \begin{align*} \int_0 ^{\infty} t \,d\mu_X(t) &= \int_0 ^\infty \left(\int_0 ^\infty {1}_{s \leq t}(s) \,ds\right) \,d\mu_X(t) \\ &= \int_0 ^\infty \left(\int_0 ^\infty {1}_{s \leq t}(t ) \,d\mu_X(t)\right) \,ds \\ &= \int_0 ^\infty \mu_X(\{ t \geq 0: s \leq t \}) \,ds \\ &= \int_0 ^\infty \mathbb{P}\{ X \in \{ t \geq 0: s \leq t \} \} \,ds \\ &= \int_0 ^\infty \mathbb{P}\{ X \geq s \} \,ds \\ &= \mathbb{E} X. \end{align*} Next fix any random variable $X$ with well-defined expectation. With $X_+ = \max(X, 0)$ and $X_- = \max(-X, 0)$, $$ \mathbb{E} X = \mathbb{E} X_+ - \mathbb{E} X_-. $$ Write $\mu_{X_+} = \mu_X|_{[0, \infty)} + \mathbb{P}\{ X < 0 \} \delta_0$. Following the non-negative case result, \begin{align*} \mathbb{E} X_+ &= \int_0 ^\infty t \,d\mu_{X_+}(t) \\ &= \int_0 ^\infty t \,d\mu_{X}(t) + \mathbb{P}\{ X < 0 \} \underbrace{\int_0 ^\infty t \,d\delta_0(t)}_{= 0} \\ &= \int_0 ^\infty t \,d\mu_X(t). \end{align*} In a similar manner, $\mathbb{E} X_- = -\int_{-\infty}^0 t \,d\mu_X(t)$...
My question is: Why is $\mathbb{E} X_- = -\int_{-\infty}^0 t \,d\mu_X(t)$ in the end? In fact, I do not think I fully understand how $$\int_0 ^\infty t \,d\mu_{X_+}(t) = \int_0 ^\infty t \,d\mu_{X}(t) + \mathbb{P}\{ X < 0 \} \int_0 ^\infty t \,d\delta_0(t)$$ for the positive case. Why does it follow from $\mu_{X_+} = \mu_X|_{[0, \infty)} + \mathbb{P}\{ X < 0 \} \delta_0$ and how do we see it for the $X_-$ case?
$\newcommand{\p}{\mathbb{P}}\newcommand{\d}{\,\mathrm{d}}$I'll explain the proof and then I'll explain there's a much more direct way to see this.
$X_+>0$ iff. $X>0$, so $\mu_{X_+}$'s restriction to $(0,\infty)$ must be $\mu_X$. $X_+=0$ iff. $X=0$ or $X<0$, so $\mu_{X_+}\{0\}=\mu_X\{0\}+\p(X<0)$. It follows that $\mu_{X_+}=\mu_X+\p(X<0)\cdot\delta_0$ is the correct law of $X_+$.
To explain: $$\int_0^\infty t\d\mu_{X_+}(t)=\int_0^\infty t\d\mu_X(t)+\p(X<0)\int_0^\infty t\d\delta_0(t)$$It suffices to explain - since, in loose notation, $\d(\p(X<0)\delta_0)=\p(X<0)\d\delta_0$ - that if $\mu,\nu$ are two measures on the same measurable space $(X,\Sigma)$ then: $$\int_Xf\d(\mu+\nu)=\int_Xf\d\mu+\int_Xf\d\nu$$For all measurable $f$.
As is often the case, it suffices to show this identity for simple functions only; in which case, it suffices to show this identity for indicator functions only. But then it's obvious: $$\int_X\chi_A\d(\mu+\nu)=(\mu+\nu)(A)=\mu(A)+\nu(A)=\int_X\chi_A\d\mu+\int_X\chi_A\d\nu$$
So we can be satisfied that, in loose notation, $\d(\mu+\nu)=\d\mu+\d\nu$ is correct.
Now, $X_->0$ iff. $X<0$, so $\mu_{X_-}$'s restriction to $(0,\infty)$ must be $\mu_{X_-}(A):=\mu_X(-A)$. $X_-=0$ iff. $X=0$ or $X>0$, so overall we find: $$\mu_{X_-}(A)\equiv\mu_X(-A)+\p(X>0)\delta_0(A)$$ For measurable $A\subseteq\Bbb R$.
Thus: $$\begin{align}\Bbb E[X_-]&=\int_0^\infty t\d\mu_{X-}(t)\\&=\int_0^\infty t\d\mu_X(-t)+\p(X>0)\int_0^\infty t\d\delta_0(t)\\&=\int_{-\infty}^0(-t)\d\mu_X(t)+0\\&=-\int_{-\infty}^0\d\mu_X(t)\end{align}$$
Here, I crucially used the pushforward-measure change-of-variables formula to replace $\d\mu_X(-t)$ with $\d\mu_X(t)$ in the way that I did. More on that later.
Overall, we get: $$\Bbb E[X]=\Bbb E[X_+]-\Bbb E[X_-]=\int_{\Bbb R}t\d\mu_X(t)$$By summing the integrals.
See the law of the unconscious statistician - a.k.a. LOTUS. This is an instance of the change-of-variables formula you get with pushforward measures. The formula goes:
Notice that $\mu_X$ is exactly $X_\ast\p$. We find: $$\int_{\Omega}g(X(\omega))\d\p(\omega)=\int_{\Bbb R}g(t)\d X_\ast\p(t)=\int_{\Bbb R}g(t)\d\mu_X(t)$$Since $X^{-1}(\Bbb R)=\Omega$. If you let $g$ be the identity function, you find exactly that: $$\Bbb E[X]=\int_{\Omega} X(\omega)\d\p(\omega)=\int_{\Omega}t\d\mu_X(t)$$
This is the most direct way. We can also use this to justify the change of variables $t\mapsto-t$ that I performed earlier. Note $A\mapsto\mu_X(-A)$ is precisely $f_\ast\mu_X$ where $f$ is the map $t\mapsto -t$. Therefore: $$\int_Ag(t)\d\mu_X(-t)=\int_Ag(t)\d f_\ast\mu_X(t)=\int_{f^{-1}(A)}(g\circ f)(t)\d\mu_X(t)=\int_{-A}g(-t)\d\mu_X(t)$$
If you let $g=\mathrm{Id}$ and $A=[0,\infty)$ you recover the identity $\int_0^\infty t\d\mu_X(-t)=\int_{-\infty}^0(-t)\d\mu_X(t)$.
Fwiw, I think $\int_0^\infty$ is poor notation when doing measure-theoretic integrals with not-necessarily absolutely continuous measures; it's ambiguous whether or not you're integrating over $[0,\infty)$ or $(0,\infty)$ and this difference may affect the value of the integral.