This is a long comment to @AnthonyQuas's solution. It works out the details of the stochastic dominance as stated by Anthony.
Let $F_n(x) = \int_{-\infty}^x \,d\mu^+_n(t)$ be the cumulative distribution function of $\mu^+_n$, and $I_n(x) = \int_{-\infty}^x F_n(t)\,d\mu^+_n(t)$. $F_n$ is a non-decreasing step function, and $I_n$ is a continuous, convex, piece-wise linear function, explicitly, $I_n(x) = \frac{1}{n}\ \sum_{k=1}^n (x - \frac{k}{n})^+$.
We want to show $I_{n+1}(x) \ge I_n(x)$ for all $x$.
For $x \le \frac{1}{n+1}$ both functions are 0.
$I_n(1) = \frac{1}{2}(1 - \frac{1}{n})$, which increases with $n$, so $I_n(1) < I_{n+1}(1)$.
Over the interval $[1,+\infty)$, $I_n(x)$ is linear, with slope 1, so for $x > 1$, $I_n(x) = (x-1) + I_n(1) < (x-1)+I_{n+1}(x) = I_{n+1}(x)$$I_n(x) = (x-1) + I_n(1) < (x-1)+I_{n+1}(1) = I_{n+1}(x)$.
Since $I_n$ is convex, and $I_{n+1}$ is piece-wise linear on the interval $[\frac{U}{n+1},\frac{U+1}{n+1}] \,\,(U=1,2,..,n)$, to show the inequality over the intervals, it suffices to show that it holds at the endpoints, i.e.: it suffices to show $I_{n+1}(\frac{U}{n+1}) \ge I_n(\frac{U}{n+1})$ for $U=1, 2, .., n+1$. The cases $U=0$$U=1$ and $U=n+1$ have been considered above.
For $U=2,.., n$, a straightforward computation yields $I_{n+1}(\frac{U}{n+1})=\frac{1}{(n+1)^2} \frac{U(U-1)}{2}$.
We have, $I_n(\frac{U}{n+1})
= \frac{1}{n}\ \sum_{k=1}^n (\frac{U}{n+1} - \frac{k}{n})^+$. Note that for the range of $U$ under consideration, and for $k=1,..,n$,
$\frac{U}{n+1} - \frac{k}{n} > 0 \iff U-k > \frac{k}{n} \iff U-k \ge 1$
therefore, $I_n(\frac{U}{n+1})
= \frac{1}{n+1}\ \sum_{k=1}^{U-1} (\frac{U}{n+1} - \frac{k}{n})=$
$=\frac{n-1}{n^2(n+1)} \frac{U(U-1)}{2} = I_{n+1}(\frac{U}{n+1})(1-\frac{1}{n^2}) $, so the inequality holds at $\frac{U}{n+1}$.