Expected value of a continuous random variable: interchanging the order of integration I have come across a proof of the following in Ross's book on Probability - For a non-negative continuous random variable Y with a probability density function $f_Y$ $$ \mathrm{E} [Y] = \int_0^\infty P[Y \geq y]dy $$ The author proves it by using $$ \int_0^\infty \int_y^\infty f_Y(x)dxdy = \int_0^\infty (\int_0^x dy) f_Y(x)dx $$ He refers to it as "interchanging the order of integration". I have studied a fair amount of Calculus from Apostol's books (Vol 1 & 2). But I still can't seem to provide a proof of this equation. How does one go about proving this last equation?

we have \begin{align*} \int_{[0,\infty)}\int_{[y,\infty)} f_Y(x)\; dx\, dy &= \int_{[0,\infty)}\int_{[0,\infty)}\chi_{[y,\infty)}(x)f_Y(x)\;dx\,dy\\\ &= \int_{[0,\infty)}\int_{[0,\infty)}\chi_{[y,\infty)}(x)f_Y(x)\;dy\,dx\\\ &= \int_{[0,\infty)}\int_{[0,\infty)}\chi_{[y,\infty)}(x)\;dy\cdot f_Y(x)\;dx\\\ &= \int_{[0,\infty)} \int_{[0,\infty)} \chi_{[0,x]}(y)\; dy\cdot f_Y(x)\; dx\\\ &= \int_{[0,\infty)} xf_Y(x)\; dx\\\ &= E(Y) \end{align*} where $\chi_A$ denotes the indicator function of a set $A$.