Convergence and continuity

Question

${f_n}$ is continuous on $\mathbb{R}$, and $f_n \to f$ uniformly on every interval $[a,b]$. Prove $f$ is continuous on $\mathbb{R}$.

I know that it must be the case that $f$ is continuous on $[a,b]$. But how can this be extended to $\mathbb{R}$?

What kind of convergence does $f_n \to f$ refer to? Because your proposition is wrong for pointwise convergence... Since you mention an interval, I'm assuming the $f_n$ are supposed to converge uniformly on every finite interval, right? — fgp, Mar 19 '14 at 00:47
The math nerd answer: Continuity is defined locally, so being continuous on any $[a,b]$ means continuous on $\mathbb R$. — Thomas Andrews, Mar 19 '14 at 01:01

score 1 · Accepted Answer · answered Mar 19 '14 at 00:51

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Choose a point $x \in \mathbb{R}$. Then $x \in [x-1, x+1]$ and $f$ is continuous on this interval, hence at $x$.

answered Mar 19 '14 at 00:51

wckronholm

And x-1, x+1 can be chosen arbitrarily? This would guarantee that the property holds for any interval? – kiwifruit Mar 19 '14 at 00:52
I'm reading your hypotheses to be that $f_n \to f$ uniformly on all intervals $[a,b]$. In particular, on the one I picked. – wckronholm Mar 19 '14 at 00:53
This would guarantee convergence on R because R is composed of closed intervals like this one? – kiwifruit Mar 19 '14 at 00:54
since $x$ is arbitrary it will be continuous on the union of all such intervals, which is $\mathbb{R}$ – Christian Chapman Mar 19 '14 at 00:55
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What I have shown is that $f$ is continuous at $x$ for all points $x \in \mathbb{R}$. This is exactly what it means for a function to be continuous on $\mathbb{R}$. – wckronholm Mar 19 '14 at 00:55

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