Example for Busemann function is not an exhaustion when Ricci $\ge 0$ For an open complete Riemannian manifold $M$ with non-negative sectional curvature, the Busemann function defined below is a convex exhaustion function (by Cheeger-Gromoll's proof of soul theorem)
The Buesemann function: 
$$b(x)=\sup_{\gamma} b_{\gamma}(x)$$ 
where the $sup$ is taken over all rays from a given point and $b_{\gamma}$ is the Busemann function associated with ray $\gamma$:
$$b_{\gamma}(x)=\lim_{t\to \infty}(t-d(x, \gamma(t))$$
A function $f:M\to \mathbb [a,\infty)$ is called an exhaustion function on $M$ if its sublevel set $\Omega_c:=f^{-1}((-\infty, c])$ is compact for all $c$ and $M=\cup_c \Omega_c$.
I am wondering that when we assume Ricci curvature is non-negative, is there any example where the Busemann function $b$ is not an exhaustion? (In this case the Busemann function is subharmonic.)
 A: This is an interesting question. I don't have an answer but I want to clarify what is being asked as there seems to be some confusion on the issue.
Given a point $p$ and a ray $\gamma$ starting at $p$ define its Busemann function $b_\gamma(x)$ by the formula $b_\gamma(x)=\lim_{t\to\infty}(t-d(x,\gamma(t)))$. This function is known to be convex if sectional curvature is nonnegative (it's an easy consequence of Toponogov comparison). Then one can define $b=\sup_\gamma b_\gamma$ where the supremum is taken over all rays starting at $p$. If $sec\ge 0$ then the sublevel sets of $b$ are convex. If a sublevel set of $b$ were noncompact it would contain a ray starting at $p$ which contradicts the definition of $b$.
This argument doesn't work for $Ric\ge 0$ because in this case the Busemann function $b$ is only known to be subharmonic rather than convex.
So the question is whether it's still true that sublevel sets of $b$ are compact for manifolds of nonnegative Ricci curvature.
Sorry for posting this as an answer but it was too long for the comment field.
A: Sorry it's not an answer. I just came across a similar question recently and found a reference. 
If the manifold has Ricci nonnegative and Euclidean volume growth, then the Busemann function is an exhaustion function. This is a result due to Zhongmin Shen.
See P400  Lemma 3.4 in Shen, Zhongmin, Complete manifolds with nonnegative Ricci curvature and large volume growth. Invent. Math. 125 (1996), no. 3, 393–404.  
But I did not check the details of this result. 
