If $\mathbb{E}\left[\lVert[ X\rVert^d\right]$ is finite then the integral in the question is necessarily finite. As mentioned, this holds whenever $\pi$ is radially decreasing. However, in the general case, you can swap regions with equal volume in $\mathbb{R}^d$ about in order to move the large probability regions closer to the origin. Doing this has no effect on the integral in question, but can only decrease $\mathbb{E}\left[\lVert[ X\rVert^d\right]$. So, it reduces to the radially decreasing case.

It isn't hard to make this idea more rigorous. If $\pi(x)$ is 'radially decreasing', so that it is a decreasing function of $\lVert x\rVert$, then

$$
\begin{align}
\iint \min\left(\pi(x),\pi(y)\right)\,dxdy&=2\iint_{\lVert y\rVert\le\lVert x\rVert}\pi(x)\,dxdy\cr
&=2K\iint\lVert x\rVert^d\pi(x)\,dx=2K\mathbb{E}\left[\lVert X\rVert^d\right]
\end{align}
$$
where $K$ is the volume of the unit ball in $\mathbb{R}^d$.
In the general case, we have
$$
\mathbb{E}\left[\lVert X\rVert^d\right]=\int_0^\infty\int_{\pi(x)\ge p}\lVert x\rVert^d\,dxdp.
$$
Letting $S_p=\lbrace x\colon\pi(x)\ge p\rbrace$ then, for a given volume $V_p$ for $S_p$, the integral $\int_{S_p}\lVert x\rVert^ddx$ is minimized when $S_p$ is a ball about the origin. In particular, if we define a radially decreasing function $\tilde\pi\colon\mathbb{R}^d\to\mathbb{R}$ by
$$
\tilde\pi(x)=\sup\lbrace p\in\mathbb{R}\colon K\lVert x\rVert^d\le V_p\rbrace
$$
then $\lbrace x\colon \tilde\pi(x)\ge p\rbrace=\lbrace x\colon K\lVert x\rVert^d\le V_p\rbrace$ has volume $V_p$. So,
$
\mathbb{E}\_{\tilde\pi}\left[\lVert X\rVert^d\right]\le\mathbb{E}\_\pi\left[\lVert X\rVert^d\right].
$

Also, the integral $\iint\min(\pi(x),\pi(y))\,dxdy$ is unchanged by passing to $\tilde\pi$. This follows from the fact that $\tilde\pi=\pi\circ f$ (almost everywhere) for some (Lebesgue) measure preserving Borel isomorphism of $\mathbb{R}^d$. Alternatively, the equality can be seen by showing that the integral only depends on $V_p$,
$$
\begin{align}
\iint \min\left(\pi(x),\pi(y)\right)\,dxdy
&=2\int\pi(x)V_{\pi(x)}\,dx\cr
&=-2\int pV_p\,dV_p=\int_0^\infty V_p^2\,dp.
\end{align}
$$

So, better than just finiteness of the integral, we have the inequality
$$
\begin{align}
\iint \min\left(\pi(x),\pi(y)\right)\,dxdy&=\iint \min\left(\tilde\pi(x),\tilde\pi(y)\right)\,dxdy\cr
&=2K\mathbb{E}\_{\tilde\pi}\left[\lVert X\rVert^d\right]\cr
&\le2K\mathbb{E}\_{\pi}\left[\lVert X\rVert^d\right],
\end{align}
$$
which is an equality whenever $\pi$ is radially decreasing.

Analysisby Lieb and Loss for more information. – Mark Meckes Sep 18 '12 at 18:56