change of variables in Lebesgue Stieljies integration

Question

Suppose a R.V has distribution function $F$. Then I believe I can write $$\int_{-\infty}^0|x|^tdF(x)=\int_{0}^{+\infty}y^tdF(-y).$$But seriously, I don't know how to rigorously justify such formula. I know this is a Lebesgue-Stieljies integral and I failed to find any good reference on change of variable theorems for it. I mean in this case, I can sort of understand why this holds, but what if I make a slightly more complicated change of variable, such as $y=x^2$. What will the formula be then? I am not even sure I can layout all the details for $y=-x$ in my case. Any pointers is greatly appreciated!

Answer 1

You can use the substitution rule ($(X,\Sigma)$ is a measurable space, $\mu,\nu$ measures on $(X,\Sigma)$) $$ \int f\,d\nu = \int f\frac{d\nu}{d\mu}\,d\mu $$ If $\lambda_\phi$ is the Lebesgue-Stieltjes measure associated to $\phi$, $$ \frac{d\lambda_\phi}{d\lambda} = \phi' $$ so you can replace $\frac{d\nu}{d\mu}$ by $\frac{d\lambda_\phi}{d\lambda} = \phi'$, and integrate w.r.t. Lebesgue measure $\lambda$.

Answer 2

Let $g(x) = -x$, $f(x) = |x|^t 1_{[0,\infty)}(x)$. Note that $\int_{-\infty}^0|x|^tdF(x)=\int f \circ g \, dF$.

You need to be a little careful about how you define $dF(-y)$.

Define a measure $\nu(A) = F(g^{-1}(A))$, where $F(A) = \int_A dF$. Hence if we let $N(\alpha) = \nu((-\infty, \alpha]) = F([-\alpha, \infty)) = 1-F((-\infty, -\alpha)) = 1 - \lim_{t \uparrow -\alpha} F(t)$ (I am assuming that $F$ is a probability measure here.)

Then the change of variables theorem (see Is there a change of variables formula for a measure theoretic integral that does not use the Lebesgue measure, for example) gives $\int f \circ g dF = \int f d N$.