= Solution
For the <exponential distribution> with mean $\mu$, $f_A(t)=F_A(t)/\mu$ and $F_A(t)=e^{-t/\mu}$. Therefore
$$
\mathbb P(T_A<T_B)=\frac1\mu\int_0^\infty F_A(t)F_B(t)\,dt
=\frac{\mathbb E X}{\mu}.
$$
Since ties have probability zero, $\mathbb P(X=T_B)=1-\mathbb E X/\mu$. Taking <expected values> in the definition of $U$ gives
$$
\boxed{\mathbb E U=\mathbb E X+\mu\mathbb P(X=T_B)
=\mathbb E X+\mu-\mathbb E X=\mu=\mathbb E T_A.}
$$
There is also an imputation interpretation. If A is observed first, $T_A=X$; if B censors A, <memorylessness of the exponential distribution> and <independence> imply that the remaining mean lifetime is $\mu$. Thus $U$ is the <conditional expectation> of $T_A$ given the observed minimum and censoring indicator. The <tower property of conditional expectation> gives the same equality. This does not say that $U$ and $T_A$ have the same distribution.
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