The uniqueness of the distribution of a random variable $\mathbf{x}$ implicitly implies that you consider a given measure $\mathbb{P}$.
Consider a probability space $(\Omega,\mathcal{F}, \mathbb{P})$, and a measurable space $(X,\mathcal{X})$. A random variable $\mathbf{x}$ is defined on $(\Omega,\mathcal{F}, \mathbb{P})$ as a measurable function $\mathbf{x}:~\Omega \to X$. Then $\mathbb{P}_{\mathbf{x}}=\mathbb{P}\circ \mathbf{x}^{-1}$ is a measure, and classically it is defined as the distribution of $\mathbf{x}$. Consider now the probability space $(\Omega,\mathcal{F}, \mathbb{Q})$, and the same function $\mathbf{x}:~\Omega \to X$. Then $\mathbb{Q}_{\mathbf{x}}=\mathbb{Q}\circ \mathbf{x}^{-1}$ is a also a measure. Therefore, if you define the random variable as a function, without a specific measure but only considering the measurable space $(\Omega,\mathcal{F})$, two different measures will give two different distributions. The KL divergence compares two measures, for a single measurable function $\mathbf{x}$.
However classically random variables are defined for a given probability measure $\mathbb{P}$, therefore have a given distribution $\mathbb{P}_{\mathbf{x}}$.
The idea is that the "dirac delta function" at $0,$ denoted by $\delta_0,$ is not an ordinary function. But somehow we can still integrate with it. For every continuous $g$ we have
$$\int_{\mathbb R} g\,\delta_0 = g(0).$$
However, no integrable function could have this property. That is, if $f$ is integrable on $[-a,a],$ then
$$\int_{-a}^a g(x) f(x)\,dx = g(0)$$
will fail for some continuous $g.$
On the other hand, and staying with elementary means, we can find a sequence $f_n$ of continuous functions on $\mathbb R$ such that
$$\lim_{n\to \infty}\int_{\mathbb R} g(x) f_n(x)\,dx = g(0)$$
for every continuous $g$ on $\mathbb R .$
Proof: Define $f_n$ on $[-1/n,1/n]$ to have the triangular graph that joins the points $(-1/n,0), (0,n), (1/n,0);$ define $f_n=0$ everywhere else. You can see that the $f_n$'s live in ever smaller intervals centered at $0,$ but nevertheless $\int_{\mathbb R} f_n = 1$ for every $n.$
Let $g$ be continuous. Then for each $n$ there exists $c_n\in [-1/n,1/n]$ such that $|g(c_n)-g(0)|$ is the maximum of $|g(x)-g(0)|$ on the interval $[-1/n,1/n].$ Thus
$$|\int_{\mathbb R} g(x) f_n(x)\,dx - g(0)| = |\int_{\mathbb R} [g(x)-g(0)] f_n(x)\,dx|$$ $$ \le |g(c_n)-g(0)| \int_{\mathbb R} f_n = |g(c_n)-g(0)|\cdot 1.$$
As $n\to \infty,$ $c_n\to 0,$ so the last expression $\to 0$ by the continuity of $g$ at $0.$
Best Answer
It means the same thing. It's just talking about sampling a number from the uniform distribution, call it X, on [-1,1], i.e. from $X\sim Unif[-1,1]$.
To calculate the covariance, you might note the formula $Cov(X,Y)=E(XY)-E(X)E(Y)$. It is known that E(X)=0, beacause X is uniform on [-1,1], so pick the middle value. $E(Y)=E(SX)=E(S)E(X)$ (beacuse they are independent), and their individual expectations are 0, so E(Y)=0. We need to find E(XY) now. Note that $E(XY)=E(SX^2)=E(S)E(X^2)$ again because S and X are independent. So we get $0\cdot E(X^2)=0$. Thus $Cov(X,Y)=0$.
To show that X and Y are dependent events, note that the value of Y is half-determined by the value of X. So Y "depends" on X. To show mathematically that they are dependent, you must find sets A and B such that $X\in A$ and $Y\in B$ are events, where $P(Y\in B|X\in A)\ne P(Y\in B)$. Choose $A=[.25,.75],B=[.25,.75]$. Then $P(Y\in B)=\frac 1 4$ (this is because Y is also uniformly distributed on [-1,1]). But $P(Y\in B|X\in A)=\frac 1 2$. This is because given that $X\in[.25,.75]$, there is equal chance that $Y\in[.25,.75]$ or $Y\in[-.75,-.25]$. Thus we have found an events such that $P(Y\in B|X\in A)\ne P(Y\in B)$, and the random variables are dependent.
Two variables are independent if $P(X\in A\cap Y\in B)=P(X\in A)P(Y\in B)$ for all possible combinations of sets A, B such that $X\in A$ and $Y\in B$. Equivalently, that $P(X\in A|Y\in B)=P(X\in A)$ for all possible combinations of events. This is obtained by dividing the first equation by $P(Y\in B)$.