Is the expectation functional continuous with respect to the sup-norm? This is a question in an Exercise of High-Dimensional Statistics by Martin J. Wainwright. 
A functional $\gamma$ is continuous in the sup-norm at $F$, if $\forall\ \epsilon>0$, there is $\delta>0$, such that $\lVert G - F \rVert_\infty < \delta$ implies that $\lvert\gamma(G)-\gamma(F)\rvert<\epsilon$, where $\lVert G - F \rVert_\infty = \sup_{t \in \mathbb{R}} \lvert G(t) - F(t) \rvert$.
The question is, suppose $F$ is a CDF, is the expectation functional $F \mapsto \int x\, dF(x)$ continuous w.r.t. the sup-norm?
 A: Well, the first problem with your hope is that not all distributions admit an expectation, so your functional isn't even defined on all CDF's. Now, be that as it may, we might hope that a uniform pertubation of a CDF which admits an expectation would then yield a CDF which also admits an expectation, but I'm afraid that this is not the case.
Consider $F(t)=1_{t\geq 0},$ that is, the distribution function of the measure $\delta_0,$ let $\delta>0$ and consider $F_{\delta}(t)=(1-\frac{1}{n}\delta)$ for $t\in [n-1,n]$ and $F_{\delta}(t)=0$ for $t<0$. Then, $||F_{\delta}-F||_{\infty}\leq \delta$ by construction. However, $F_{\delta}$ does not admit an expectation.
To see this, let $X_{\delta}$ be distributed according to $F_{\delta}$ and note that, since $X_{\delta}$ is a positive variable, we can apply Tonelli's theorem to get that
$$
\mathbb{E}(X_{\delta})=\mathbb{E}\int_0^{X_\delta} 1\textrm{d}t=\mathbb{E}\int_0^{\infty}1_{X_{\delta}> t}\textrm{d}t=\int_0^{\infty} \mathbb{E}1_{X_{\delta}> t}\textrm{d}t=\int_0^{\infty}1-F_{\delta}(t)\textrm{d}t =\sum_{n=1}^{\infty} \frac{\delta}{n}=\infty
$$
Hence, there is not even an open subset of the space of CDF's that admits an expectation, and you can modify $F_{\delta}$ to see that there are CDF's uniformly close to $F$ which admit arbitrarily large expectations. Thus, no, the expectation is not a uniformly continuous functional on any natural subspace of CDF's.
