Let $p_1$ and $p_0$ be two probability density functions and let the function defined by $$l(y)=\frac{p_1(y)}{p_0(y)}$$ be the probability ratio function.
Random variable $Y$ has a density function $p_0$ and the following integral is calculated:
$$\Theta=\int_{\{y:\,l(y)\leq t\}} p_1(y)\mathrm{d}y=\int_{\{y:\,l(y)\leq t\}} \frac{p_1(y)}{p_0(y)}p_0(y)\mathrm{d}y=\int_{\{y:\,l(y)\leq t\}} l(y)p_0(y)\mathrm{d}y=E_{P_0}[l(Y)|l\leq t]$$
I read in a course material that if one makes the the change of variable as $x=l(y)$, then the same integral can be written as
$$\Theta=\int_{0}^t x p_{l}(x)\mathrm{d}x=E_{P_l}[l|l\leq t]$$
where $p_l$ is the density of the random variable $l(Y)$. Remember that $Y$ is distributed as $P_0$, which accepts the density $p_0$.
The question is how to show that indeed $$E_{P_0}[l(Y)|l\leq t]=E_{P_l}[l|l\leq t]$$
Supplementary information (if necessary):
I tried to show this with no success. What I know is the following:
$$\Theta^{'}=\int_{\{y:\,l(y)\geq t\}} p_0(y)\mathrm{d}y=\int_{t}^\infty p_l(x)\mathrm{d}x$$
where again $p_l$ is the density of $l(Y)$, when $Y\sim P_0$. In a similar way and now assuming that $p_l$ is the density of $l(Y)$, when $Y\sim P_1$, one can also write
$$\Theta^{''}=\int_{\{y:\,l(y)\leq t\}} p_1(y)\mathrm{d}y=\int_{0}^t p_l(x)\mathrm{d}x$$
We need $p_l$ to be such a function that the change of variable transformation is: $$p_0(y)= p_l(l(y))\;\left\lvert \frac{\operatorname d l(y)}{\operatorname d y}\right\rvert$$
So $$\begin{align} \int_{\{y:0\leq l(y)\leq t\}} l(y)\,p_0(y)\operatorname d y & = \int_{\{y|0\leq l(y)\leq t\}} l(y)\; p_l(l(y))\;\left\lvert \frac{\operatorname d l(y)}{\operatorname d y}\right\rvert\;\operatorname d y \\ & = \int_0^t x\; p_l(x)\operatorname d x \end{align}$$