Question: Is there a single symbol that represents the set of $x\pm 1$ and $x$?

Thus, if $x$ is 8, I'm looking for the most concise way to present the set of $\{7,8,9\}$.

I apologize if this has been asked before; there are many questions on the plus-minus sign, but I couldn't find my answer elsewhere.

Answer: 
Taking integration by parts to the [*Dirichlet integral*](https://en.wikipedia.org/wiki/Dirichlet_integral), it is easy to check that

$$ \int\_{-\infty}^{\infty} \frac{1-\cos(at)}{t^2} \, dt = \pi|a|. \tag{1}$$

Taking advantage of the fact that the integrand of $\text{(1)}$ is non-negative, by the [*Tonelli's theorem*](https://en.wikipedia.org/wiki/Fubini%27s_theorem#Tonelli.27s_theorem_for_non-negative_functions), for *any* real-valued random variable $Z$ we have

$$ \pi \Bbb{E}[|Z|] = \Bbb{E}\left[ \int\_{-\infty}^{\infty} \frac{1-\cos(Zt)}{t^2} \, dt \right] = \int\_{-\infty}^{\infty} \frac{1-\Bbb{E}[\cos(Zt)]}{t^2} \, dt. $$

Therefore

\begin{align\*}
\pi \Bbb{E}[|X+Y| - |X-Y|]
&= \int\_{-\infty}^{\infty} \frac{\Bbb{E}[\cos((X-Y)t)-\cos((X+Y)t)]}{t^2} \, dt \\
&= \int\_{-\infty}^{\infty} \frac{\Bbb{E}[2\sin(Xt)\sin(Yt)]}{t^2} \, dt \\
&= \int\_{-\infty}^{\infty} \frac{2\Bbb{E}[\sin(Xt)]^2}{t^2} \, dt \\
&\geq 0.
\end{align\*}

Moreover, notice that the equality holds if and only if $\Bbb{E}[\sin(Xt)] = 0$ for all $t$. This means that the c.f. $\varphi\_X(t) = \Bbb{E}[e^{itX}]$ is real-valued, which is equivalent to the symmetry condition: $X \stackrel{d}{=} -X$.
Here's another argument. It doesn't seem to generalize to $p$ norms, but perhaps it is instructive anyway.

For independent variables $X,Y$ with finite expectation we can write

\begin{align\*}
\mathbb E|X-Y|
&=\int\_{-\infty}^\infty \mathbb P[X\leq t< Y]+\mathbb P[Y\leq t<X] dt\\
&=\int\_{-\infty}^\infty F\_X(t)(1-F\_Y(t))+F\_Y(t)(1-F\_X(t)) dt\tag{1}
\end{align\*}

where $F\_Z$ denotes the cumulative distribution function $F\_Z(t)=\mathbb P[Z\leq t]$ of a random variable $Z.$

In the case at hand, $X$ and $Y$ are i.i.d., so $F\_{-Y}(t)=1-F\_X(-t).$ Using $-Y$ instead of $Y$ in (1) gives

$$\mathbb E|X+Y|=\int\_{-\infty}^\infty F\_X(t)F\_X(-t)+(1-F\_X(-t))(1-F\_X(t)) dt\tag{2}$$

We can get a comparable integrand for $\mathbb E|X-Y|$ by substituting $t$ for $-t$ in the final term in (1), and using $F\_Y=F\_X$:
$$
\mathbb E|X-Y|=\int\_{-\infty}^\infty F\_X(t)(1-F\_X(t))+F\_X(-t)(1-F\_X(-t)) dt\tag{3}
$$

Writing $a=F\_X(t)$ and $b=F\_X(-t),$ clearly $((1-a)-b)((1-b)-a)=(1-a-b)^2\geq 0,$
so $a(1-a)+b(1-b)\leq ab+(1-a)(1-b).$ Integrating over $t$ and applying (3) and (2) gives $\mathbb E|X-Y|\leq \mathbb E|X+Y|.$ The equality case is when $F\_X(-t)+F\_X(t)=1$ a.e., which is when $X$ is symmetric about zero.