Showing $\int_\Omega f(h(w))dP=\int_S f(y)d(P\circ h^{-1})$

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Why does the following hold ?

If $h:\Omega\to S$ then; $\displaystyle\int_\Omega f(h(w))dP=\int_S f(y)d(P\circ h^{-1})$

I know the change of variable formula:

$\displaystyle\int_{h(a)}^{h(b)}f(x)dx=\int_a^b f(h(t))h'(t)dt$

but in my case $h$ appears in the 'delta part', can I write the RHS as;

$\displaystyle\int_S f(y)d(P\circ h^{-1})=\int_{h(\Omega)} f(y)dP\circ(h^{-1})dh^{-1}=\int_\Omega f(h(w))\left(dP\circ(h^{-1}\circ h\right) \left(dh^{-1}\cdot h'\right)$ Can one consider $dh^{-1}$ as $\left(h^{-1}\right)$' ?

Of course one has to request that $h$ is continuously differentiable or so, but my questions is only; how is the change of variable made here, thanks.