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Let $Y$ be a subset of $[0,1]^{[0,1]}$ of integrable functions. $I: Y \rightarrow \mathbb{R}$, $f \mapsto \int_{0}^{1}f$. We use the product topology on $[0,1]^{[0,1]}$, that is, we consider $f$ as $\prod_{x \in [0,1]}{f(x)}$ and the euclidean metric topology on $\mathbb{R}$. How do we prove that $I$ is not continuous?

Now for all $f_n$ in $Y$ with $f_n \rightarrow f$ we have $I(f_n) \rightarrow I(f)$ since $f_n$ converges to $f$ uniformly, but I know this doesn't necessarily mean that $I$ is continuous.

Moishe Kohan
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hteica
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    It really isn't true that $f_n\to f$ uniformly. The product topology can equivalently be thought of as the topology of pointwise convergence.

    That said, since you've restricted to positive and bounded functions, convergence is immediate by the dominated convergence theorem.

     – FShrike Apr 23 '23 at 11:55
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    https://math.stackexchange.com/a/54068/8157. (One of my favorite answers of the present website.) – Giuseppe Negro Apr 23 '23 at 12:00
  • @FShrike why? If we choose some open neighborhood around $f$ then we’ll find that $(f_n)$ will eventually be in it, doesn’t this simply mean that the supremum of the absolute difference goes to zero? – hteica Apr 23 '23 at 12:10
  • @GiuseppeNegro could you elaborate about the function being surjective on open sets and how the statement follows from that? – hteica Apr 23 '23 at 12:14
  • @hteica No it doesn't. Neighbourhoods of $f$ are merely products of neighbourhoods of $f(x_i)$ for finitely many $x_i$. Nothing can be said about the suprema – FShrike Apr 23 '23 at 12:15
  • @FShrike ah I forgot about it being only for finitely many $x_i$, but correct me if I’m wrong: if we allow the neighborhoods of $f$ to be products of neighborhoods of $f(x_i)$ for infinitely many $x_i$ then $f_i$ should converge to $f$ uniformly, right? – hteica Apr 23 '23 at 12:18
  • But that's not a neighbourhood in the product topology. Such a set: $$\prod_{x\in[0,1]}U_x$$Where $U_x$ is a proper open subset of $[0,1]$ for all $x$, is not open at any point in the product topology – FShrike Apr 23 '23 at 12:20
  • @FShrike yes I know, I was just wondering. Thanks :) – hteica Apr 23 '23 at 12:20

1 Answers1

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The convergence in the product topology is not uniform, but pointwise.

For sequences, $I$ will be continuous by Dominated Convergence. But let $\def\cF{\mathcal F}\cF$ be $$\cF=\{F\subset[0,1]:\ \text{finite}\},$$ ordered by inclusion.

Let $$ f_F(x)=\begin{cases}1,&\ x\in F\\[0.3cm] 0,&\ \text{otherwise}\end{cases} $$ Then $f_F\to1$ but $I(f_F)=0$ for all $f_F$ and $I(1)=1$.

Edit: how to "avoid" nets. Continuity by nets versus continuity by neighbourhoods are just two sides of the same coin. Here, you can consider the function $f(x)=1$; to say that $I$ is not continuous at $f$ means that there exists $\varepsilon>0$ (which we can take to be $1$ in this case) such that for every neighbourhood $N$ of $f$ there exists $f_N\in N$ such that $|I(f)-I(f_N)|≥1$. Given a neighbourhood $N$ of $f$ there exists a basic neighborhood $f\in N_F\subset N$ for some $F\in\cF$ and $\delta>0$, where $$ N_0=\{g:\ |f(x)-g(x)|<\delta,\ x\in F\}. $$ Now we can take $f_N=f_F$, so $f_N\in N$ and $I(f)-I(f_N)=1$.

Martin Argerami
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    But the functions should be $[0,1] \to [0,1]$, no? So it seems that the integral indeed is continuous, by the Lebesgue dominated convergence theorem. – mechanodroid Apr 23 '23 at 11:58
  • @mechanodroid I don't think the integral is continuous, since its part of an exercise to show that it is not continuous – hteica Apr 23 '23 at 11:58
  • My bad, I missed the codomain. The problem is more subtle, then as $I$ is sequentially continuous for measurable functions by dominated convergence. – Martin Argerami Apr 23 '23 at 12:00
  • I have changed the answer. – Martin Argerami Apr 23 '23 at 12:06
  • This is a nice way to show $[0,1]^{[0,1]}$ is not a sequential space, exhibiting sequential but not netwise continuity. – FShrike Apr 23 '23 at 12:14
  • So it is not even sequential continuous? Its interesting because in the book that I’m reading its remarked that $I$ is an example of a sequential continuous function but not continuous – hteica Apr 23 '23 at 12:20
  • @hteica No, it is sequentially continuous in the example of a subspace of $[0,1]^{[0,1]}$ – FShrike Apr 23 '23 at 12:21
  • It is sequentially continuous. It's a direct consequence of Dominated Convergence, since your functions are measurable and bounded. – Martin Argerami Apr 23 '23 at 12:21
  • @MartinArgerami yes, thank you – hteica Apr 23 '23 at 12:22
  • Do you have any example that can show that $I$ is not continuous? – hteica Apr 23 '23 at 12:25
  • Yes, it's what my answer is about. – Martin Argerami Apr 23 '23 at 12:26
  • @MartinArgerami sorry I may be very slow but didn’t you just show that there’s a sequence $(f_i)$ such that $f_i \rightarrow f$ but $I(f_i)$ doesn’t converge to $I(f)$? – hteica Apr 23 '23 at 12:28
  • Net, not sequence; but yes. Which precisely shows that $I$ is not continuous. – Martin Argerami Apr 23 '23 at 12:33
  • I understand now, you’re using filter. I’m wondering if there’s any examples which do not use filter? Thanks in adv – hteica Apr 23 '23 at 12:34
  • As said in my answer and again in the comments, your $I$ is sequentially continuous. – Martin Argerami Apr 23 '23 at 12:35
  • No I'm saying if we can prove that $I$ is not continuous without using any filter. – hteica Apr 23 '23 at 12:38
  • Sorry, I cannot make sense of what you are saying. If you only use sequences, $I$ is continuous. But $I$ is not continuous because continuity fails for nets. To show that continuity fails for nets you will have to use nets. Unless what you mean is that you want to use neighbourhoods? – Martin Argerami Apr 23 '23 at 12:48
  • Sorry I was just wondering if its possible to prove that I is not continuous without using any notions of filter/nets :) – hteica Apr 23 '23 at 14:18
  • "Avoiding nets" can be done by cumbersomely writing the same in the language of neighbourhoods. I have edited that in. – Martin Argerami Apr 23 '23 at 14:37