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Prove that if $F:(−\infty,a]\rightarrow\mathbb{R}$ and $G:(a,\infty]\rightarrow\mathbb{R}$ are both continuous and $F(a) = G(a)$ then the function $H$ defined by $$ H(x) = \begin{cases} F(x) & \text{if $x\leq a$}\\ G(x) & \text{if $x\geq a$}\\ \end{cases} $$ is also continuous

My attempt:

Let $a\in\mathbb{R}$, consider the sequence of intervals $[a, a + \frac{1}{n}]$ for all $n\in\mathbb{N}$. Select a rational number $x_n$ from each interval and we will get a rational sequence $\{x_n\}$ which converges to a. Then $F(x_n) = G(x_n)$ for all $n\in\mathbb{N}$. Also by the continuity of $F(x)$ and $G(x)$, we have $\lim_{n\rightarrow\infty}F(x_n)=\lim_{n\rightarrow a}F(x)=F(a)$ and $\lim_{n\rightarrow\infty}G(x_n)=\lim_{n\rightarrow a}G(x)=G(a)$. Therefore, $F(a)=G(a)$, for all $a\in\mathbb{R}$.

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    $\begingroup$ I don't understand your attempt at all. To be more precise: why do you choose the intervals? Why does it matter that $x_n$ are rational? Why $f(x_n)=g(x_n)$ (also you mean $F,G$ not $f,g$)? $\endgroup$ – Yanko Oct 21 '18 at 20:49
  • $\begingroup$ I think you mean the intervals $[a, a + 1/n]$ instead. Although even with the correction, your proof doesn't work. You know that $F$ and $G$ are equal only at $a$. szw1710 gives a good argument to use, or at worst go back to the $\epsilon$-$\delta$ definition. $\endgroup$ – bitesizebo Oct 21 '18 at 20:51
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It seems to me it is not needed to be so precise. We could say that one-sided limits at $a$ are equal to each other. Indeed, $H(a-)=F(a-)=F(a)=G(a)=G(a+)=H(a+)$. Of course $H(a)=H(a-)=H(a+)$. This gives continuity at $a$. $H$ is trivially continuous at any $x_0\ne a$.

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