Bounded monotone convergence for functions Let $f:[a,b]\rightarrow\mathbb{R}$ be a bounded, increasing function. Prove that $\;\lim\limits_{x\to b^-} f(x)=\sup\limits_{x\in\left[a,b\right[}f(x)\;.$
This looks a lot like the principle of bounded monotone convergence for sequences, but instead for limits at a point as opposed to limits at infinity. I tried using the sequential characterisation of limits but got nowhere (as we must take a limit to infinity)
 A: This is not true in general:
$$f : [0,1] \rightarrow \Bbb R$$
$$f(x) := \begin{cases} x & \text{ if } x \in [0,1)\\
2 & \text{ if } x = 1
\end{cases}$$
$f$ is increasing but $\lim_{x \to 1}f(x) = 1 < 2 = \sup_{[0,1]}f$. (Note that an increasing function on a compact interval is always bounded as $f(a) \leq f(x) \leq f(b)$ for every $x \in [a,b]$)
However the claim holds if $f : [a,b] \rightarrow \Bbb R$ is continuous at $b$:
$$\lim_{x \to b}f(x) = f(b) = \sup_{[a,b]}f =: M$$
The last equality follows from the fact that $f$ is increasing, indeed for every $x \in [a,b]$ $f(x) \leq M$ by definition, thus
$$f(b) \leq M$$
but also $f(x) \leq f(b)$ for every $x \in [a,b]$, so
$$M \leq f(b)$$
It is interesting to note that an increasing funtion $f : [a,b] \rightarrow \Bbb R$ has at most a countable number of discontinuity points, so the requirement of continuity at $b$ makes sense.
Addendum: proof with $\epsilon$-$\delta$ argument, let $l := \lim_{x \to b}f(x)$, by definition
$$\forall \epsilon > 0 \: \exists \delta > 0 \: \text{s.t.}\: x \in (b-\delta,b) \Rightarrow |l - f(x)| < \epsilon$$
now
$$l = l - f(x) + f(x) < \epsilon + M \Rightarrow l \leq M$$
(up to now we have not used continuity)
In the next step the contiuity is needed
$$\forall \epsilon > 0 \: \exists \eta > 0 \: \text{s.t.}\: x \in (b-\eta,b] \Rightarrow |f(b) - f(x)| < \epsilon$$
Now let $\alpha := \min\{\delta,\eta\}$, then for $x \in (b - \alpha,b)$ it holds:
$$|f(b) - f(x)| < \epsilon$$
$$|l - f(x)| < \epsilon$$
finally
$$M = f(b) = f(b) - f(x) + f(x) -l + l < 2 \epsilon + l \Rightarrow M \leq l$$
For $f$ not continuous at $b$ one might be tempted to argue as follow: by definition of $\sup$
$$\forall \epsilon > 0 \: \exists x_\epsilon \in [a,b] \: \text{s.t.}\: M - f(x_\epsilon) < \epsilon \tag{1}$$
then $M - f(x) < \epsilon$ for all $x \in [x_\epsilon,b]$ by monotonicity. Define
$$c:= \max\{x_\epsilon,b - \delta\} \tag{2}$$
then for all $x \in (c,b)$
$$M - f(x) < \epsilon$$
$$|l - f(x)| < \epsilon$$
so
$$M = M - f(x) + f(x) - l + l < 2 \epsilon + l \Rightarrow M \leq l$$
however if $f$ is discontinuous at $b$ then $l < M$ so taking any $\epsilon < M - l$ in $(1)$ forces $x_\epsilon = b$ and the condition $(2)$ gives $c = b$ therefore $(c,b) = (b,b) = \emptyset$ and the argument does not work.
As said in the comment the following equality holds:
$$\lim_{x \to b} f(x) = \sup_{[a,b)}f =: L$$
if the supremum is achieved say at $x_0$ then $f(x) = L$ for every $x \in [x_0,b)$ and the equality holds, so let's assume that the supremum is not achieved. Arguing as before it is immediate to prove $l \leq L$. Now
$$\forall \epsilon > 0 \: \exists x_\epsilon \in [a,b) \: \text{s.t.}\: L - f(x_\epsilon) < \epsilon$$
Let $c$ as above, then for all $x \in (c,b)$
$$L - f(x) < \epsilon$$
$$|l - f(x)| < \epsilon$$
so
$$L = L - f(x) + f(x) - l + l < 2 \epsilon + l \Rightarrow L \leq l$$
in this case $(c,b)$ is always non-empty
