I was reading a probability course, i have difficulties to understand one line of a proof.
X1's CDF is F(We don't know his law).
U1 follows an Uniform distribution U[0,t].
$ P(U_{1} + X_{1} \geq t) = \frac{1}{t} \int_{0}^{t} P(U_{1} + X_{1} \geq t | U_{1} = s)ds = \frac{1}{t} \int_{0}^{t} (1-F(u))du. $
Any ideas of how we obtain $\frac{1}{t} \int_{0}^{t} (1-F(u))du$ in the end?
Thanks in advance!
By conditioning on $U_1$ we compute \begin{align} \mathbb P(U_1+X_1\geqslant t) &= \int_{\mathbb R} \mathbb P(U_1+X_1\geqslant t\mid U_1=s)f_{U_1}(s)\ \mathsf ds\\ &= \int_0^t\frac1t \mathbb P(U_1+X_1\geqslant t\mid U_1=s)\ \mathsf ds\\ &= \frac1t\int_0^t \mathbb P(X_1\geqslant t-s)\ \mathsf ds\\ &= \frac1t \int_0^t \mathbb P(X_1\geqslant u)\ \mathsf du\\ &= \frac1t \int_0^t \left(1-\lim_{v\to u^-}F(v)\right)\ \mathsf du. \end{align} Note that $X_1$ was not assumed to be a continuous random variable, so $F$ need not be continuous - hence the use of the left limit of $F$.