= Solution
If $y_1\leq y_2$, then the <utility function> is increasing, so for every $X\in\mathcal X$,
$$
\mathbb E[U(X+y_1)]\leq\mathbb E[U(X+y_2)].
$$
Taking <suprema> gives $F(y_1)\leq F(y_2)$.
For $0\leq t\leq1$, let $X_1$ and $X_2$ attain the suprema at $y_1$ and $y_2$. Because $\mathcal X$ is a <vector space>,
$$
X_t=tX_1+(1-t)X_2\in\mathcal X.
$$
The <concavity> of $U$ now gives
$$
\begin{aligned}
F(ty_1+(1-t)y_2)
&\geq\mathbb E\left[U\left(X_t+ty_1+(1-t)y_2\right)\right]\\
&\geq t\mathbb E[U(X_1+y_1)]
+(1-t)\mathbb E[U(X_2+y_2)]\\
&=tF(y_1)+(1-t)F(y_2).
\end{aligned}
$$
\b[Thus $F$ is increasing and concave, as stated by <optimized affine shift of concave utility>.]
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