Find $\mathbb{E}(h(X) \mid U)$ where $h$ is measurable, $X, Y$ are independent and $U = \max(X,Y)$ 
Find $\mathbb{E}(h(X) \mid U)$ where $h$ is measurable, $X, Y$ are independent and $U = \max(X,Y)$.
  $X$, $Y$ follow both an exponential distribution with parameter $\lambda = 1$.

I couldn't find a way to solve this problem as $(X,U)$ doesn't have a probability density function in $\mathbb{R^2}$.
Any other ways to proceed ?
 A: As an alternative to the method mentioned in the @Math1000's comment (and a related post) you may compute this conditional expectation as follows (assuming that $h(X)$ is integrable):
$$
\mathsf{E}(h(X)\mid U=u)=\lim_{\delta\downarrow 0}\frac{\mathsf{E}[h(X)1\{U\in [u,u+\delta)\}]}{\mathsf{P}(U\in [u,u+\delta))}.
$$
For example, assuming that $h$ is continuous,
$$
\mathsf{E}(h(X)\mid U=u)=\frac{1}{2}h(u)+\frac{1}{2}\int_0^u \frac{h(x)e^{-x}}{1-e^{-u}}dx
$$
because
\begin{align}
\mathsf{E}[h(X)1\{U\in [u,u+\delta)\}]&=(e^{-u}-e^{-(u+\delta)})\int_0^u h(x)e^{-x}\,dx \\
&\quad+(1-e^{-(u+\delta)})\int_u^{u+\delta}h(x)e^{-x}\,dx,
\end{align}
and
\begin{align}
\mathsf{P}(U\in [u,u+\delta))&=(1-e^{-(u+\delta)})^2-(1-e^{-u})^2.
\end{align}
The result follows from applying L'Hôpital's rule and taking the limit ($\delta\downarrow 0$).
A: Clearly, 
$$ E[ h(X) g(U)] = \underset{=(*)}{\underbrace{E[h(X)g(X),X>Y]}}+ \underset{=(**)}{\underbrace{E[h(X) g(Y),X<Y]}}.$$
Now 
\begin{align*}  &(*) = \int_0^\infty  g(u) h(u) e^{-u} (1-e^{-u})du\mbox{ and }\\ 
&(**) = \int_0^\infty g(u) e^{-u} H(u)du,
\end{align*} 
where $$\boxed{H(u) = \int_0^u e^{-x} h(x) dx}$$ 
Thus, 
$$ E[h(X) g(U) ] =\int_0^\infty g(u) e^{-u}  [ h(u)(1-e^{-u})  + H(u)] du.$$
Pause and consider the particular case $h\equiv 1$, in which  $H= 1-e^{-u}$. With this choice, the equation above shows that  $U$ has density $f_U(u) = 2e^{-u} (1-e^{-u})$. Armed with this information, let's return to the general case: 
\begin{align*}  E [ h(X) g(U) ] &= \int_0^\infty g(u) f_U(u)\times \frac 12 \left (h (u) + \frac{H(u)}{1-e^{-u}}\right) du\\
& = E[ g(U) \frac 12 \left(h(U) + \frac{H(U)}{1-e^{-U}}\right)]
\end{align*}
The answer is therefore 
$$ \boxed{E[ h(X) | U ] = \frac 12 \left( h(U) + \frac{H(U)}{1-e^{-U}}\right)}$$
Note that the method can be used for more general cases. 
