Consider two distributions
$$F(x) =\begin{cases} 0 &\text{if} \ \ x < 0\\ \frac{1}{4} &\text{if} \ \ 0\leq x < 2\\ 1 &\text{if} \ \ 2\leq x \end{cases}$$ and $$G(x) =\begin{cases} 0 &\text{if} \ \ x < 1\\ 1 &\text{if} \ \ 1\leq x \\ \end{cases}$$ Show that the expected value of $x$ under $F$ is greater than the expected value of $x$ under $G$ but that $F$ does not first-order stochastically dominate $G$.
Attemped solution - It seems the expected value of both distributions will be infinity, so I do not know how to show the latter. For first-order stochastic dominance we have $$U(F) = \int_{-\infty}^{0}dx + \int_{0}^{2}\frac{1}{4}dx + \int_{2}^{\infty}dx$$ again it does not seem that we will get any meaningful evaluation.
I'll give you the details that you need for this and have you fill in any details (or ask me, and I can put them in). Let $\mathbb{E}_{F}[X]$ be the expected value of $X$ under $F$, and similarly for $\mathbb{E}_{G}[X]$.
Notice that $F$ is a distribution function and not a probability mass/density function - so you would not be integrating or summing $F$, but integrating or summing the probability density or probability mass function, respectively. This applies similarly to $G$.
$F$ is constant throughout, so that means there are point masses. The probability mass function corresponding to $F$ consists of a probability mass of $\dfrac{1}{4}$ at $x = 0$ and a probability mass of $\dfrac{3}{4}$ at $x = 2$. Hence, $$\mathbb{E}_{F}[X] = \dfrac{1}{4}(0)+\dfrac{3}{4}(2)=\dfrac{6}{4}=1.5\text{.}$$
Similarly for $G$, there is a probability mass of $1$ at $x = 1$, hence $$\mathbb{E}_{G}[X] = 1(1) = 1\text{.}$$ Hence, $\mathbb{E}_{F}[X] > \mathbb{E}_{G}[X]$.
$F$ does not first-order stochastically dominate $G$. If this were the case, then $F(x) \leq G(x)$ for all $x$ with strict inequality at some $x$. However, for $x \in [0, 1)$, $F(x) = \dfrac{1}{4}$ and $G(x) = 0$ so $F(x) > G(x)$.