If $f$ is monotone increasing on an interval and has a jump discontinuity at $x_0$, show that the jump is bounded above by $f(x_1)-f(x_2)$

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If $f$ is monotone increasing on an interval and has a jump discontinuity at $x_0$ in the interior of the domain show that the jump is bounded above by $f(x_1) - f(x_2)$ for any two points $x_1$, $x_2$ of the domain surrounding $x_0$, $x_1 < x_0 < x_2$.

So I've I tried solving this here is what I have:

Let $f$ be monotone increasing on an interval $A$ that has a jump discontinuity at $x_0$ on the interior of the domain where $x_0 \in A$. Let there be any $x_1, x_2 \in A$ where $x_1< x_0 < x_2$. Then by definition of monotone increasing $f(x_1) \leq f (x_2)$.

From here I want to say that $f(x_1) \leq f(x_0) \leq f(x_2) \rightarrow f(x_1) - f(x_1) \leq f(x_0) \leq f(x_2) - f(x_1)$ and that $f(x_0) \leq f(x_2) -f(x_2)$. But I'm not sure if I'm going in the right direction since I can't say for sure that $f(x_0)$ is really less than $f(x_2)-f(x_1)$. Or would I need to do something like $f(x_0)-f(x_1)\leq f(x_2)-f(x_1)$ and go from there. Any help would be appreciated as I'm somewhat unsure on this problem!

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$x_1$ and $x_2$ are supposed to be given points. You cannot let $x_2 \to x_1$. If $x_0<t<x_2$ then $f(t) \leq f(x_2)$. Let $t \to x_0$ to get $f(x_0+) \leq f(x_2)$. [ $f(x_0+)$ is the right hand limit of $f$ at $x_0$]. Similarly prove that $f(x_0-) \geq f(x_1)$. From these two se get $f(x_0+)-f(x_0-) \leq f(x_2)-f(x_1)$.

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First of all you have to define " the jump at $x_0$ "

It means $$\lim _{x\to x_0+}f(x) -\lim _{x\to x_0-}f(x) $$

The above limits exist due to the monotonicity of $f(x)$

The rest is not too complicated to prove and you can do it.