This might be an obvious question, but I couldn't prove it rigorously.
Suppose $f(x)$ is continuous at every $a$. Then, for every a for every $e>0$, there exists $d>0$ such that $(|x-a|<d \Rightarrow \left|f(x)-f(a)\right|<e).$
Let $g(x)=g(x+t)$. I want to show that for every $a$ for every $e>0$, there exists $d>0$ such that $(|x-a|<d \Rightarrow \left|g(x)-g(a)\right|<e).$
\begin{align*} |g(x)-g(a)| &=|f(x+t)-f(a+t)|\\ &=|f(x+t)-f(x)+f(x)-f(a+t)| \\ & \leq |f(x+t)-f(x)|+|f(x)-f(a+t)|\\ &=|f(x+t)-f(x)|+|f(x)-f(a)+f(a)-f(a+t)| \\ & \leq |f(x+t)-f(x)|+|f(x)-f(a)|+|f(a)-f(a+t)|. \end{align*}
But I dont know how to go further. I've spent so much time on this simple problem, but still cant figure it out.
Thanks!