This is a question from my math book:

Let $a< b < c$. Suppose that $f$ is continuous on $[a,b]$, $g$ is continuous on $[b,c]$, and $f(b) = g(b)$. Define $h$ on $[a,c]$ by $h(x) = f(x)$ for $x\in [a,b]$ and $h(x) = g(x)$ for $x \in [b,c]$. Prove that $h$ is continuous on $[a,c]$.

What I want to do is prove that $h$ is continuous on $[a,c]$ but not at $b$.

I'm thinking that I have to pick an $α$ such that $a<α< b$, and then show that $h(x)$ is continuous at $α$. So basically, I want to show that $∀\epsilon>0, ∃ δ>0$ such that if $x\in [a,c]$ and $|x-α|<δ$ then $|h(x)-h(α)|<\epsilon$. And from the given information, I know that $f:[a,b]\to \mathbb{R}$, and $∀ \epsilon>0, ∃ δ>0$ such that if $x\in [a,b]$ and $|x-α|<δ_f$ then $|f(x)-f(α)|<\epsilon$.

How do I connect these two definitions to find $δ$? And is there anything else I need to prove? Like are there any cases I should be making?

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What you want to show is false; $h$ will be continuous at $b$. For points in $[a,c]$ that are actually in $[a,b)$, you can use the continuity of $f$ directly, taking care not to "go" past $b$; for points that are in $(b,c]$, you can use the continuity of $g$ (again, taking care not to go "past" $b$); for the point $b$, you'll need to combine the fact that $f$ is continuous at $b$ (remember that this means that $f$ is continuous from the left at $b$) and that $g$ is continuous at $b$ from the right. – Arturo Magidin Mar 25 '12 at 4:07

What you have to prove is precisely that $h$ is continuous at $b$. Because at any point $d$ other than $b$, if $d<b$ then $h(x)=f(x)$ for all $x$ in a small interval around $d$, and so the continuity of $f$ implies the continuity of $h$ for every $d\in[a,b)$. Similarly one deduces that $h$ is continuous on $(b,c]$ by using the continuity of $g$.

For the continuity at $b$. Fix $\varepsilon>0$. By the continuity of $f$ at $b$, there exists $\delta_1>0$ such that if $x<b$ and $b-x<\delta_1$, then $|f(x)-f(b)|<\varepsilon$. Similarly, there exists $\delta_2>0$ such that if $x>b$ and $x-b<\delta_2$, then $|g(x)-g(b)|<\varepsilon$.

Let $\delta=\min\{\delta_1,\delta_2\}$. Now, if $|x-b|<\delta$, we consider two cases: first, if $x<b$, then $b-x=|x-b|<\delta\leq\delta_1$, and so $$|h(x)-h(b)|=|f(x)-f(b)|<\varepsilon;$$ if $x>b$, then $x-b=|x-b|<\delta\leq\delta_2$, and so $$|h(x)-h(b)|=|g(x)-g(b)|<\varepsilon.$$ So $h$ is continuous at $b$.

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We can prove that $h$ is continuous on $[a,c]$ in three steps:

First, we prove that $h$ is continuous on $[a,b)$. Since $f$ is continuous on $[a,b)$, given any $x\in [a,b)$ and $\epsilon>0$ we have some $\delta>0$ such that for $y\in [a,b)$ we have $|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$. Let $\delta'=\min\{\delta,b-x\}$. Since $h=f$ on $[a,b)$, we have that for $y\in [a,c]$, $|x-y|<\delta'$ implies that $y\in [a,b)$ and $|x-y|<\delta$ so $|h(x)-h(y)|<\epsilon$, thus $h$ is continuous at $x$. Hence $h$ is continuous on $[a,b)$.

Second, we prove that $h$ is continuous on $(b,c]$. This proof is nearly identical to the previous one.

Finally, we prove that $h$ is continuous at $b$. Since $f$ and $g$ are continuous at $b$, given any $\epsilon>0$ we have some $\delta,\delta'>0$ such that for $y\in [a,b]$ if $|b-y|<\delta$ then $|f(b)-f(y)|<\epsilon$ and for $y\in [b,c]$ if $|b-y|<\delta'$ then $|g(b)-g(y)|<\epsilon$. Thus we can let $\delta''=\min\{\delta,\delta'\}$ and we get that if $y\in [a,c]$ and $|b-y|<\delta''$ then either $y\in [a,b]$ and $|b-y|<\delta$ in which case $|h(b)-h(y)|=|f(b)-f(y)|<\epsilon$, or $y\in [b,c]$ and $|b-y|<\delta'$ in which case $|h(b)-h(y)|=|g(b)-g(y)|<\epsilon$. Thus $h$ is continuous at $b$.

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