Mean residual life If X is a nonnegative random variable representing the life of a component having distribution function F,the mean residual life is defined by
$$
m(t) = E(X-t | X >t) = \frac{1}{\bar F(t)} \int_t^\infty (x-t) d\nu(x),       t\geq 0
$$
In papaer R. C. Gupta and D. M. Bradley (2003)" Representing the Mean Residual Life
in Terms of the Failure Rate"mentioned that by writing $$x - t = \int_{t}^{x} du$$  and employing Tonelli's theorem yields the equivalent formula
$$
m(t) = \frac{1}{\bar F(t)}\int_t^\infty \int_t^x du d\nu(x) = \frac{1}{\bar F(t)}\int_t^\infty \int_u^\infty d\nu(x) du = \frac{1}{\bar F(t)}\int_t^\infty  \bar{F}(u) du
$$
How can we get this result by substituting the above integral and using Tonelli's theorem?
 A: Here is another way you can think about the problem.
\begin{align*} 
& \mathbb{E}[X - t \, | \, X> t] = \int_0^{\infty} \mathbb{P}(X - t > s  \, | \, X > t)ds = \int_0^{\infty} \dfrac{\mathbb{P}(X - t > s, \, X > t)}{\mathbb{P}(X > t)}ds \\
= & \dfrac{1}{\mathbb{P}(X > t)}\int_0^{\infty} \mathbb{P}(X > t + s, \, X > t)ds = \dfrac{1}{1 - F(t)} \int_0^{\infty} (1 - F(t+s))ds \\= & \dfrac{1}{1 - F(t)} \int_t^{\infty} (1 - F(s))ds
\end{align*}
A: You have a nonnegative random variable $X\newcommand{\E}{\mathbb{E}}$ with distribution function $F$. The mean residual life is defined as
$$
  m(t) = \E\left[X - t\mid X >t\right] = \frac{\E\left[ (X-t) I_{\{X>t\}}\right]}{P\{X>t\}} =  \frac{1}{1-F(t)} \int_t^\infty (x-t)\,dF(x) \, ,
$$
for $t>0$. But 
$$
  \int_t^\infty (x-t)\,dF(x) = \int_t^\infty \left(\int_t^x du\right)dF(x) \, . \quad (*)
$$
Tonellis' Theorem says that you can change the order of integration in $(*)$, but you have to be careful about the integration limits. Look at the following figure.

The original domain of integration is interpreted like this: $x$ varies from $t$ to $\infty$. For some fixed $x$, $u$ varies from $t$ to $x$, determining the filled region in the figure. Now, reverse the order: $u$ varies from $t$ to $\infty$. For a fixed $u$, $x$ varies from $u$ to $\infty$. Hence,
$$
  \int_t^\infty \left(\int_t^x du\right)dF(x) =  \int_t^\infty \left(\int_u^\infty dF(x)\right)du 
$$
$$
  = \int_t^\infty P\{X > u\}\,du = \int_t^\infty \left(1 - F(u)\right)\,du \, .
$$
Therefore,
$$
  m(t) = \frac{1}{1-F(t)} \int_t^\infty \left(1 - F(u)\right)\,du \, .
$$
