A very close questions has been already posted and migrated there math.stackexchange.com/questions/4996629/…

Do you want to find one submatrix of all submatrices $a \times m$ with smallest rank? Is $a$ fixed too?

Where does this inequality comes from?

dense' should be replaced twice with total', isn't it?

If you do want to get an answer, you should correct your post and the notations, making explicit everywhere the dependence of $\nu$ with regard to $p$. For example what does $\mu_{p_i} \to \nu$ mean? Who is $\nu$? I clearly wish to down vote the question is no effort of clarification in made.

The Lipschitzness of the cumulative distribution function is a simple sufficient condition for absolute continuity. Yet, as I wrote, it is a too strong condition.

So you should correct and clarify the post. Note that you got two opposite answers, probably because the question is not clear. Are you convicted by the arguments given?

I do not understand DCT. Any Lipschitz function is absolutely continuous.

Do not set $\nu:=\nu_p$ if this measure depends on $p$, otherwise what is the meaning of $\mu_{p}:=\frac{1}{p}\sum_{i=0}^{p-1}T_{\ast}^{i} \nu$? Do you mean $\mu_{p}:=\frac{1}{p}\sum_{i=0}^{p-1}T_{\ast}^{i} \nu_p$? Be explicit to avoid confusions and loss of time!

It is still not clear. An example of upper semi-continuous linear map is $F \mapsto \mu(F)$ where $F$ is a closed set. For arbitrary bounded measurable $\phi$, the map $\mu \mapsto \int\phi\d\mu$ may be neither upper semi-continuous nor lower semi-continuous. Morevoer, $\phi$ $\mu$-a.e. bounded does not make sense since $\phi$ should be fixed whereas $\mu$ varies. I will edit a bit my answer.

Your edit is not correct. What does $\mu_p \to \mu_p$ and $f(\mu_p) \to (\mu_p)$ when $p \to \infty$ mean? Was not $p$ fixed when you introduced the measure $\nu$? Be clear in your questions if you wish answers.

A more known result is that $p$ divides $\binom{p}{k}$ whenever $1 \le k \le p-1$. The congruence you give is a simple consequence (by recursion on $k$), thanks to the recursion relation satisfied by binomial coefficients. So Pascal probably knew it.