Does pointwise convergence of a sequence of functions in each closed subinterval of an open interval imply pointwise convergence in the open interval?

Question

If a sequence {$f_n$} be uniformly convergent in every closed subinterval $[a+ε,b-δ]⊂(a,b)$, then it isn't necessarily uniformly convergent in $(a,b)$. But what about pointwise convergence? That is, if a sequence {$f_n$} be pointwise convergent in every closed subinterval $[a+ε,b-δ]⊂(a,b)$, is it pointwise convergent in $(a,b)$? {$f_n$} may or may not be uniformly convergent.

Answer 1

Of course it is. Pick a point $x \in (a,b)$. The interval being open gives you an $ \epsilon>0 $ such that $(x-\epsilon, x+\epsilon) \in (a,b)$, and clearly this open interval contains the closed interval $[x-\frac{\epsilon}{2},x+\frac{\epsilon}{2}]$.

Answer 2

Assume $f_n$ is not pointwise convergent. Then there is some point $x_0\in (a,b)$ where the function values $f_n(x_0)$ do not converge as $n\to\infty$. Can you see the contradiction?

Answer 3

For each point $x$ in this interval the set $\{x\}$ is a closed subinterval.