# Convergence of a sequence of functions: pointwise, uniform, and almost everywhere

I'm trying to reason things out when it comes to convergence of functions and I haven't found an answer that confirms or dismisses these two points:

• A sequence of functions $\{f_n\}$ could converge a.e. to a function $f$, either pointwise or uniformly, depending on $f$ and the choice of its domain.

• The same arguments for a.e. convergence can be made when $n$ is not a natural number, but a real number.

Are these claims correct?

• Both assertions are correct. – Evan Aad Sep 12 '16 at 22:13
• I'm not sure I follow the second claim. You are indexing a sequence of function $\{f_n\}$ by the natural numbers so that $$\{f_n\} = \{f_1, f_2, f_3, \dots\}$$ The sequence is always countable, so we might as well label it with $\mathbb{N}$ being our index set. You could label the sequence with real numbers, but the index set, would still have to be countable, meaning you could convert this to the conventional style of indexing by the naturals. You normally do not index by an uncountable index set for a sequence of functions, so how does $n$ being a real number come into play? – J. Marx-Kuo Sep 12 '16 at 22:27
• @J.Marx-Kuo take the set of functions $\{f_\epsilon\}$ where $f_\epsilon=\exp(-x^2/\epsilon ^2)$ – ToniAz Sep 12 '16 at 23:27
• Ok, so this is more a matter of terminology. I would argue that $\{f_{\epsilon}\}$ is a family of functions, as opposed to a sequence. Sequences are always enumerated (meaning countable), but I don't want to harp on this because it's not what you're asking about. To get back to the question, I would say the second claim is true, in that a (potentially uncountable) family of functions can converge pointwise, uniformly, etc. depending on the function and the domain. – J. Marx-Kuo Sep 13 '16 at 0:10