Continuity issue for convergence to functional of Brownian motion

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Let $e_{t} \sim IID(0,1)$ and set $k=[T\pi]$ for $\pi \in [0,1]$. We know from the functional CLT that $\sum_{t=1}^{k}e_{t}/\sqrt{T} \Rightarrow B(\pi)$. I am now interested in the quantity $Z_{T}(k)=\sqrt{\frac{T^{2}}{k(T-k)}} T^{-1/2}\sum_{t=1}^{k}e_{t}$ which should converge in distribution to $B(\pi)/\sqrt{\pi(1-\pi)}$. My understanding is that for this to hold we need continuity of the functions $\pi \mapsto 1/\sqrt{\pi}$ and $\pi \mapsto 1/\sqrt{1-\pi}$ and therefore the convergence result should hold solely for $\pi \in (0,1)$.

Now suppose I want to establish the distribution of $\sum_{k=1}^{T}Z_{T}(k)/T$. Can I now appeal to the continuous mapping theorem to argue that this latter quantity converges in distribution to $\int_{0}^{1} (W(s)/\sqrt{s(1-s)}) ds$? It seems to me that the argument would be incorrect due to the continuity issue at the 0 and 1 boundaries. Yet the limiting object seems to be well defined since $\int_{0}^{1}1/\sqrt{s(1-s)}<\infty$. Could anyone help pinpoint where the problem with this argument is?

J