# Clarification on a Claim in Black-Scholes-Merton Derivation

In these notes: https://johnthickstun.com/docs/bscrr.pdf, towards the end of the proof of Proposition 5.2 on page 6, the author claims:

$$\log \lim_{n \to \infty} \Bbb{E}_\pi \left[\frac {S^*_n} S \right] = \log \Bbb{E}_\nu \left[\frac {S^*} S \right]$$

I.e., that

$$\lim_{n \to \infty} \mathbb{E}_\pi \left[\frac {S^*_n} S \right] = \mathbb{E}_\nu \left[\frac {S^*} S \right]$$

I'm not entirely following why that's true. In Proposition 4.2, the author does prove that

$$\log \frac {S^*_n} S \xrightarrow [n \to \infty] {d} \mathcal{N}(\nu T, \sigma^2 T)$$

in distribution (weakly), but that's weaker than convergence in mean, which is what he seems to be using. He does refer to Proposition 4.1, but Proposition 4.1 talks about the log-returns, not the returns, and in any case, doesn't speak to the limit. So even if I try to compute the expected value of $$\frac {S^*_n} S$$ with respect to the risk-neutral probability measure $$\pi$$ directly, it ends up being $$\mathbb{E}_\pi \left[\frac {S^*_n} S \right] = T \log r$$, which he already has. What am I missing?