Limit everywhere, limit function is continuous, specific proof. Suppose $f:[a,b] \to R$ is a function such that $\lim_{t\to x} f(t) = g(x)$ exists $\forall x \in [a,b]$.
It can be shown that $g(x)$ is a continuous function. I seem to remember that there was a proof using Riemann Integrability.
Did anyone come across such a proof? If so, what was it?
 A: I have tried working this out, and have come up with a proof:  
Let $\epsilon’ > 0$. Let $\epsilon = \frac{\epsilon’}{4} > 0$.  As $lim_{t \to x} f(t) = g(x)$, we have that $\exists \delta > 0$ such that if $0 < |t - x| < \delta$, then $|f(t) - g(x)| < \epsilon$. First note that if $t,t_2$ are such that $0 < |t - x| < \delta$ and $0 < |t_2 - x| < \delta$, then $$|f(t) - f(t_2)| < 2\epsilon$$
Now, we have already established that $\exists \delta > 0$ such that if $|t - x| < \delta$, then $|f(t) - g(x)| < \epsilon$. Also, considering a point $t$ where $0 < |t - x| < \delta$, we know that $\lim_{t_2 \to t} f(t_2) = g(t)$. Thus, we have that $\exists \delta_t > 0$ such that if $0 < |t - t_2| < \delta_2$ then $|f(t_2) - g(t)| < \epsilon$. Let $\delta’_t = min(\delta_t, \frac{min((|t - (x + \delta)|, |t - (x - \delta)|)}{2})$. If $0 < |t_2 - t| < \delta’_t$, then note that $0 < |t_2 - x| < \delta$ and thus $|f(t) - f(t_2)| < 2\epsilon$. We also have that, for a $t$ such that $0 < |t-x| < \delta$, if $|t_2 - t| < \delta’_t$, then $|f(t_2) - g(t)| < \epsilon$ and $|f(t) - g(x)| < \epsilon$. 
Thus, we have that $|f(t_2) - g(t)| + |f(t) - g(x)| < 2 \epsilon$. Thus,   $|f(t_2) - f(t) + g(x) - g(t)| < 2 \epsilon$. This means that $ -2\epsilon < f(t_2) - f(t) - g(t) + g(x) < 2 \epsilon$. As $|f(t) - f(t_2)| < 2\epsilon$, we have that, for $0 < |t-x| < \delta$, $|g(t) - g(x)| < 4 \epsilon = \epsilon’$. 
Thus, for any $\epsilon’ > 0$, $\exists \delta > 0$, such that if $0 < |t - x| < \delta$, then $|g(t) - g(x)| < \epsilon’$. Thus, $g$ is continuous on $[a,b]$.
Please correct me if my proof has any logical flaws. My proof does not use Riemann integration though. Although I can imagine how such a proof will be possible, since the interval is closed and $g$ is continuous, it will be uniformly continuous. 
