TITLE: Ricci curvature under rough convergence
QUESTION [5 upvotes]: From the work of Lott--Villani and Sturm, I know that the following fact holds:

(*) Suppose that $(M_k,g_k,dvol_{g_k})$ is a sequence of compact Riemannian manifolds of non-negative Ricci curvature that converges in the measured Gromov--Hausdorff sense to $(M_\infty,g_\infty,dvol_{g_\infty})$, which is a smooth Riemannian manifold. Then $(M_\infty,g_\infty)$ also has non-negative Ricci curvature. 

The proof I have in mind goes like: $(M_k,g_k,dvol_{g_k})$ has non-negative Ricci curvature in the weak optimal transport sense. This is preserved under measured Gromov--Hausdorff convergence. Hence $(M_\infty,g_\infty,dvol_{g_\infty})$ has non-negative Ricci curvature in the optimal transport sense. Because it is smooth, this then implies that it has non-negative Ricci curvature in the usual sense. 
I think, however, that this fact was known long before the optimal transport interpretation of Ricci lower bounds (there were lots of works on GH convergence and lower Ricci bounds, e.g. Cheeger and Colding had a sequence of papers studying the regularity of such limits (the first one is here). However, I could not find a reference for this. So, my question is:

Can (*) be proven without the optimal transport theory?

EDIT: It seems like there was some ambiguity in my question, as I did not specify if the sequence was allowed to collapse to a lower dimension. I had the case of non-collapsing in mind, but am very interested in the general case as well. So, to summarize, I believe that the non-collapsing case (i.e. $M_k$ and $M_\infty$ have the same dimension), there was a well known proof, while in the collapsed case, optimal transport provided the first proof.

REPLY [5 votes]: Let me try to prove it.
Let $M$ be an $m$-dimensional Riemannian manifold.
Set 
$$f_p(z)=\frac{|p-z|_M^2}2,$$ 
where $|p-z|_M$ denotes the distance from $p$ to $z$ in $M$.
Note that $M$ has nonnegative Ricci curvature if and only if 
$$(\Delta f_p)(x)\le m$$
for any $p$ and $x\in M$.
The later means that for the integral
$$\int\limits_{B_r(x)}f_p(z)\cdot d_z\mathrm{vol}$$
the comparison inequality holds, i.e., if $\tilde p,\tilde x\in\mathbb{E}^m$ and $f_{\tilde p}(\tilde x)=f_{p}(x)$ then
$$\int\limits_{B_r(x)\subset M}f_p(z)\cdot d_z\mathrm{vol}
\le 
\int\limits_{B_r(\tilde x)\subset \mathbb{E}^m}f_{\tilde p}(\tilde z)\cdot d_{\tilde z}\mathrm{vol}.$$
The last inequality survives in measured Gromov--Hausdorff limit.