Show that $\lim_{x\to a^{+}} g(x) =g(a)$ 
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a bounded function and suppose that $g(x)=\sup_{t>x}f(t)$. Show that $\lim_{x\to a^{+}}g(x) = g(a)$ for all real $a$.

I have a hard time with these kind of proofs because in high school, I would simply replace $x$ by $a$ in $g(x)$ to get that $g(x) \rightarrow g(a)$. Obviously, Real Analysis requires a more formal proof.
I can always apply the definition of a limit of a function :
Given $\varepsilon>0$, there exists $\delta>0$ such that if $x\in A$ and $0<|x-c|<\delta$, then $|f(x)-L| < \varepsilon$.
In particular, we have that if
$$ |x-c| = |x-a^{+}|<\delta$$
then
$$|g(x)-g(a)| < \varepsilon$$
How can I find a relation between $f$ and $g$, the sup of $f(t)$ and between $x$ and $a^{+}$?
 A: It should be clear that $g$ is non-increasing. If $x<y$, then since $(x,\infty) \supset (y,\infty)$ we have $g(x) \ge g(y)$.
Now suppose $x_n \downarrow a$. We have $g(x_n) \uparrow \gamma$, and $g(x_n) \le \gamma \le g(a)$ (since $x_n > a$).
Suppose $\gamma < g(a)$, and let $\epsilon= \frac{1}{2}(g(a)-\gamma)>0$. Then there is some $x>a$ such that $g(x) > g(a)-\epsilon$ (by definition of $\sup$), and so for some $n$ we have $a \le x_n \le x$. Since $g$ is non-increasing we have $g(x_n) \ge g(x) > g(a)-\epsilon = \frac{1}{2}(\gamma+g(a)) \ge \gamma$, which is a contradiction. Hence $\gamma = g(a)$.
A: Here is a useful way to think about supremums (suprema?):

If $S$ is a nonempty set, then if $\sup S = M$ exists, then for every $\epsilon
> 0$ there is some $s \in S$ so that $M - \epsilon < s$.

Now if I hand you some $a$ and $\epsilon > 0$, you know by definition that there has to be some $x > a$ so that $g(a) - \epsilon < f(x)$, by the above fact. So then, by definition of $g$, you know $f(x) \leq g(x)$.
Now show that $g$ is decreasing and combine the above facts.
