is $S(n,k) = S^{*}(n,k)$? Egoroff's Theorem: If $\mu(X)<\infty$, if $\{f_n\}$ is a sequence of complex measurable functions which converges pointwise on $X$, and if $\epsilon >0$, then there is some measurable $E\subset X$ with $\mu(E-X)<\epsilon$ such that $\{f_n\}$ converges uniformly on $E$.
I'm studying the proof of Egoroff's theorem and I have seen that the definition of the auxiliary set has different versions. In one of the tests they defined it as follows:
$$S(n,k) = \bigcap_{i>n}\{x \in X \colon |f_{i}(x)-f(x)|<\frac{1}{k} \}$$
On the other hand, in another test they define it as follows:
$$S^{*}(n,k) = \{x \in X \colon |f_{i}(x)-f(x)|<\frac{1}{k},\hspace{0.2cm} \text{for all} \hspace{0.2cm} i > n \}$$
My question is, are these two sets the same?
On the other hand, in the proof of this theorem, where is the fact that $ \mu(X) <\infty $ used?
Proof of Egoroff's theorem:

Also, I would like to know how to show that the theorem does extend, with essentially the same proof, to the situation in which the sequence $\{f_n\}$ is replaced by a family $\{f_t\}$, where $t$ ranges over the positive reals; the assumptions are now that, for all $x\in X$,
(i) $\hspace{0.2cm}$ $\lim_{t \to \infty}f_{t}(x) = f(x)$ and
(ii) $\hspace{0.2cm}$ $t \to f_{t}(x)$ is continuous
Any answer is appreciated and valued
 A: The two sets $S(n,k)$ and $S^*(n,k)$ are the same : this is a phenomena where the intersection can be taken inside the set-notation as a qualifier.
Let me give you another example. Let's say your universal set consists of different kinds of lollipops, and you're in a crowd of children, each holding some lollipops. You want to describe the set of lollipops that are held by children in the crowd.
On the one hand, you look at each child, what lollipop they are holding, and then put it together. That would lead to :
$$
\bigcup_{\text{child $\in$ crowd}} \{\text{lollipops held by child}\}
$$
On the other hand, you could just say : tell me all the lollipops so that at least one child has that kind of lollipop. That particular request would translate to the equal set $$
\{\text{lollipop : child holds lollipop for at least one child $\in$ crowd }\}
$$
You can replace the text "at least one" by "some" as well. This is all done pretty much to match mathematical notation to intuition : if you can't understand the intersection or union well, then switch to the textual definition, and vice-versa.
An example of this is illustrated in measure-theory, when you translate logical statements like the definition of continuity (for all $\epsilon>0$, there is $\delta>0$ etc.) which are text-based definitions, into rigorous set-theoretic union-intersection notation. So being able to switch between the notations is important.
$S(n,k)$ is more illustrative for mathematical purposes (understood as an infinite intersection over some sets) : it would make more sense to work with $S(n,k)$ compared to $S^*(n,k)$ if you're writing an exam, for example.

As for the other question about $\mu(X)<\infty$ : up till the point $\mu(X) = \lim_{n \to \infty}\mu(S(n,k))$, nothing about $\mu(X)$ is used (the limit follows from the monotone convergence theorem, which doesn't require finiteness).
However, following this line, the "thus , for each $k$, there is some $N_k$ such that " requires $\mu(X) < \infty$. Indeed, the point is that $\mu(X) = \lim_{n \to \infty}\mu(S(n,k))$ doesn't imply anything about $\mu(X - S(n,k))$ unless $\mu(X) < \infty$, because if the sequence $\mu(S(n,k))$ converged to infinity, there could be no control over the quantity $\mu(X - S(n,k))$. For example, consider $X = \mathbb R$ with the Lebesgue measure, and $S(n,k) = [-n,n]$. Then, even though $\mu(X) = \lim_{n \to \infty}\mu(S(n,k))$ is true, $\mu(X - S(n,k))$ is infinite each time.
On the other hand, if $\mu(X)$ is finite then we know that $\mu(X - S(n,k)) = \mu(X) - \mu(S(n,k))$ as an equality over the real numbers (i.e. involving no infinities), and therefore it is true by limit laws that $\lim_{n \to \infty} \mu(X - S(n,k)) = 0$. That's exactly what allows the existence of $N_k$ such that $\mu(X - S(n,k))  < \frac{\epsilon}{2^k}$ for $n \geq N_k$.

We have $S(n,k)=\cap_{i > n}\{x \in X : |f_i(x) - f(x)| < \frac 1k\}$.
Interpretation
Remember what you're trying to do. You're trying to make $f_n$ uniformly converge on some set. Now, a sequence of functions is uniformly convergent on a set $X$ if and only if it's uniformly Cauchy on that set.
What is uniformly Cauchy? In words : a sequence of functions $f_n$ is uniformly convergent on a set $F$ if , for all $\epsilon>0$, there exists $N$ such that $n,m>N$ implies $|f_n(x) - f_m(x)|<\epsilon$ for all $x \in F$.
Let's try to understand what kind of set is useful for profiling the uniform Cauchy property. Let's imagine we do this : for each $\epsilon$ and $N$, we find a good set of $x$, where $f_n$ satisfies the Cauchy definition at that point , at least for the given $\epsilon$ with choice $N$. Once we take the intersection over all $\epsilon >0$ and $N$ large enough, we expect to get a set of points on which $f_n$ behaves in a uniformly Cauchy fashion, because we've filtered the good points out at every stage, so a point that's good enough at every stage will be a good point to be included in a set where $f_n$ uniformly converges.
So $S(n,k)$, which you can think of as $S(n,\frac 1k)$ (so that $\frac 1k$ is taking on the role of $\epsilon$) is the set of all $x$ for which the sequence $f_n(x)$ satisfies the definition of a Cauchy sequence with $\epsilon = \frac 1k$ and $N = n$. Basically, $S(n,k)$ is a collection of candidate points $x$ on which we get $f_n(x)$ to satisfy the definition for
at least $\epsilon = \frac 1k$ and $N = n$.
Further interpretation
The eventual set of uniform convergence, is what? Look back at the definition of uniformly Cauchy : for all $\epsilon >0$, there is an $N\ in \mathbb N$ such that $i,j >N$ implies $|f_i(x) - f_j(x)|<\epsilon$. As you may know, the $\epsilon$ can be replaced by a countable sequence , so one may use the equivalent definition :

for all $k\in \mathbb N$, there is an $N\in \mathbb N$ such that $i,j >N$ implies $|f_i(x) - f_j(x)|<\frac 1k$

Let's write this in set-theoretic notation now. "For all" becomes an intersection (an element belongs to the intersection of some sets, if and only if it belongs to all of them), "there is" becomes a union (an element belongs to the union of some sets if and only if it belongs to at least one of them) , and what about the rest? The "such that $i,j$..." That's $S(n,k)$!
In other words, a good candidate set for uniform convergence of the sequence $f_n$, is the set described by $$
E' = \cap_{k \in \mathbb N} \cup_{N \in \mathbb N} S(N,k)
$$
Now, you obviously see what is happening. The important thing is that $S(N,k)$ is an increasing sequence in $N$ i.e. $S(1,k) \subset S(2,k) \subset S(3,k) \subset ...$, we know that $S(N_k,k) = \cup_{n=1}^{N_k} S(n,k)$. The set $E$, which is in the proof above, is nothing but $$
E = \cap_{k=1}^\infty \cup_{n=1}^{N_k} S(n,k)
$$
which, if you notice, is actually a very careful stripping out from $E$, to ensure that it is really close to $E'$, but not quite there. That's the point : the difference between the two sets, by choice of $N_k$, is at most $\epsilon$. This sort of clears the air around $S(n,k)$ and uniform convergence, and there's also the example I gave above, of translating "qualifiers" to "set-theoretic notation" which you will do well to take on!
