Suppose $f:[0,1]\rar[0,1]$ is measurable and $X_1,X_2,\ldots$ an i.i.d. sequence in $[0,1]$ with distribution $\mu$. Define
$$
N\colon=\inf\{2n:\,X_{2n-1}\leq f(X_{2n})\}
$$
then
$$
\P(X_N>t)
=\int_t^\infty\mu((0,f(x)])\,\mu(dx)/
\int_0^\infty\mu((0,f(x)])\,\mu(dx)~.
$$
If $\mu=\l$, then $X_N$ has density $f/\int f\,d\l$.
Since $[N=2n]=[X_{2n-1}\leq X_{2n}]\cap\bigcap_{j=1}^{n-1}[X_{2j-1}>X_{2j}]$, we get by independence:
\begin{eqnarray*}
\P(X_N>t)
&=&\sum_{n=1}^\infty\P(X_{2n}>t,X_{2n-1}\leq f(X_{2n}))
\prod_{j=1}^{n-1}\P(X_{2j-1}>f(X_{2j}))\\
&=&\sum_{n=1}^\infty\P(X_{2n}>t,X_{2n-1}\leq f(X_{2n}))
\P(X_{1}>f(X_{2}))^{n-1}\\
&=&\P(X_{2}>t,X_{1}\leq f(X_{2}))/
(1-\P(X_{1}>f(X_{2})))
\end{eqnarray*}
Thus the assertion follows from
$$
\P(X_{2}>t,X_{1}\leq f(X_{2}))=
\int_t^\infty\mu((0,f(x)])\,\mu(dx)~.
$$