Given $2^n$ samples of $r_i$ drawn from the distribution $\nu_0$ (assume i.i.d.). Also, I know that $\nu_0(r=ess\inf r)>0$. Let $$ g(\alpha)=-\alpha\ln\left(E_{\nu_0}[e^{-\frac{r}{\alpha}}]\right)\qquad \text{and}\qquad g_{2^n}(\alpha)=-\alpha\ln\left(\frac{1}{2^n}\sum_{i=1}^{2^n}e^{-\frac{r_i}{\alpha}}\right). $$ Suppose $\alpha^*=\arg\max_{\alpha\geq 0}g(\alpha)=0$ and we let $\alpha_{2^n}^*=\arg\max_{\alpha\geq 0}g_{2^n}(\alpha)$.
Now, I consider the following two events $$ G := \{\exists r_i=ess\inf r\}\cap\{\forall r_i\geq ess\inf r\}\qquad \text{and}\qquad Z := \{\alpha^*_{2^n}=0\} $$
I am trying to prove that $E\left[|g_{2^n}(\alpha_{2^n}^*)-g(\alpha^*)|\mathbb{1}_{G\cap Z}\right]=0$.
What I did was trying to show that $g_{2^n}(\alpha_{2^n}^*)-g(\alpha^*)=0$, when $\alpha^*_{2^n}\to 0$ and $\alpha \to 0$. By trying some numerical examples, I find that $\lim_{\alpha\to 0}g_{2^n}(\alpha)=1$ (not rigourous), but I could not tell anything about $\lim_{\alpha\to 0}g(\alpha)$. Otherwise, we can also try to show that $P(G\cap Z)=0$, which I think is more difficult. I might need some intuition into solving this. Thanks!