I am studying an exercise about Egorov's theorem and the last question is about a generalization to a continuous sequence. More precisely, let $(X,\mu)$ be a measure space of finite measure, and $\{f_t\}$ be a continuous sequence of complex functions indexed by positive real numbers satisfying:
(i) $\forall x\in X, f_t(x)\rightarrow_{t\rightarrow\infty} f(x)$
(ii) $\forall x\in X, t\rightarrow f_t(x)$ is continuous.
I don't see why the exercise supposes assumption (ii). Indeed, defining the sets
$$S(t,k)=\cap_{j\geq t}\left\{x\in X: |f_j(x)-f(x)|\leq \frac{1}{k}\right\}$$
for $k$ positive integer, we have $\mu(S(t,k))\rightarrow \mu(X)$ when $t\rightarrow\infty$ thanks to assumption (i) and the fact that $\mu(X)<\infty$, so that we can define an increasing sequence $\{t_k\}$ satisfying $\mu(S(t_k,k))\geq \mu(X)-\frac{\epsilon}{2^k}$. Considering $E=\cap_{k>0} S(t_k,k)$, we have $\mu(X - E)\leq \sum_{k=1}^{\infty} \frac{\epsilon}{2^k}=2\epsilon$ and uniform convergence of $\{f_t\}$ on $E$.
Is this proof false or is assumption (ii) not necessary?
As it stands, the proof is not wrong but rather incomplete: this proves that $\sup_{x\in E}|f_{t_k}(x)-f(x)|\to 0$ as $k$ goes to infinity, but not necessarily as $t\to\infty$. If in the definition of $S(t,k)$ we take the intersection over the real numbers $j\geqslant t$, we may loss the measurability.
But we can define $$S(t,k):=\bigcap_{\substack{s\geqslant t\\s\in\mathbb Q }}\{x\in X,|f_s(x)-f(x)|\leqslant k^{-1}\},$$ then use a continuity argument.