I see, thank you for clarifying! So the missing step is to show that $\|(AB_n)^+-B_n^+A^+\|$ tends to zero (in probability?) as $n$ goes to infinity, and this should follow from the fact that $B_n$ is almost orthogonal as $n$ tends to infinity. Am I correct?

Oh I see, sorry for the silly question. One last comment: which type of convergence are you considering when you write $\lim_{n\to \infty} \frac{1}{n} B_n B_n^*=\sigma^2 I$?

Thanks for your answer. Could you please elaborate a little more (or provide a reference) on the derivation of the first two displayed equations?

@FedericoPoloni: Thanks for your comment! I was wondering whether the rows of $B$ need to be orthonormal for $(AB)^+=B^+ A^+$ to hold true.

@N.T.: Yes, I would say that $B_n$ is of full (row or column, depending on $m$) rank with probability 1 for every $n$. Why?