Property of intervals of invisible points I know Riesz's lemma stating that

For any function $f:[a,b]\to\mathbb{R}$ having only step discontinuities the set of the points invisible from the right (and, respectively, from the left) is open in $[a,b]$ and therefore is the union of countably many open disjoint intervals $(a_k,b_k)$ (where one of the set can be of the form $[a,b')$),

where a point $x_0\in[a,b]$ is defined as invisible from the right for a function $f:[a,b]\to\mathbb{R}$ when $\exists\xi\in(x_0,b]:$ $\max\{f(x_0^+),f(x_0),f(x_0^-)\}<f(\xi)$.
Kolmogorov-Fomin's Introductory Real Analysis states that, for such intervals, the following equality holds:$$f(a_k^+)\leq\max\{f(b_k^+),f(b_k),f(b_k^-)\}$$Could anybody explain why that inequality is true? I suppose that the English translation Introductory Real Analysis does not contain an error here, although the original Russian text has one.
Moreover, if we define a point $x_0\in[a,b]$ as invisible from the left for a function $f:[a,b]\to\mathbb{R}$ when $\exists\xi\in[a,x_0):$ $\max\{f(x_0^+),f(x_0),f(x_0^-)\}<f(\xi)$, is some analogous inequality true on the respective $(a_k,b_k)$ intervals? I have thought for example about the possibility that $f(a_k^-)\geq\max\{f(b_k^+),f(b_k),f(b_k^-)\}$ might hold. I heartily thank you!
EDIT: inserted $\max$ thanks to David's remark.
 A: Assumptions


*

*That by "step discontinuities" you mean that $f(x+)$ and $f(x-)$ exist at every point of discontinuity.

*That you want the inequality: $f(a_k+)\leq\max\{f(b_k-),f(b_k),f(b_k+)\}$ (your post leaves out the "max").

*I can use Riesz's Lemma.


Preliminaries
The interval $(a_k,b_k)$ contains only points that are invisible from the right. If $x\in(a_k,b_k)$, then $$(1)\text{ }\max\{f(x+),f(x),f(x-)\}<f(\xi_x)\text{ for some } \xi_x>x.$$ 
Since the intervals $(a_k,b_k)$ are disjoint and their union consists of all points invisible from the right, $b_k$ is not invisible from the right. Thus $$(2)\text{ }\max\{f(b_k+),f(b_k),f(b_k-)\}\geq f(\xi)\text{ for all }\xi>b_k.$$
Furthermore, suppose that $x\in(a_k,b_k)$ and there is $\xi_x< b_k$. Let $\xi_x^{\prime}$ be the supremum of such $\xi_x$. 
If $\xi_x^{\prime}<b_k$ then $\xi_x^{\prime}$ is invisible from the right. If $\xi_{\xi_x^{\prime}}<b_k$, then $$\max\{f(x+),f(x),f(x-)\}\leq f(\xi_x^{\prime})\leq\max\{f(\xi_x^{\prime}+),f(\xi_x^{\prime}),f(\xi_x^{\prime}-)\}<f(\xi_{\xi_x^{\prime}})$$ but this shows that $\xi_{\xi_x^{\prime}}$ is one of the $\xi_x$, but it is larger than their supremum. This contradiction establishes that $\xi_{\xi_x^{\prime}}\geq b_k$. Consequently, $$\max\{f(x+),f(x),f(x-)\}\leq\max\{f(\xi_x^{\prime}+),f(\xi_x^{\prime}),f(\xi_x^{\prime}-)\}\\
<f(\xi_{\xi_x^{\prime}})\leq\max\{f(b_k+),f(b_k),f(b_k-)\}.$$
On the other hand, if $\xi_x^{\prime}\geq b_k$, then (2) gives us $$\max\{f(x+),f(x),f(x-)\}\leq f(\xi_x^{\prime})\leq\max\{f(b_k+),f(b_k),f(b_k-)\}$$
In either case, we obtain $$(3)\text{ }\max\{f(x+),f(x),f(x-)\}\leq\max\{f(b_k+),f(b_k),f(b_k-)\}\\
\text{ for all }x\in(a_k,b_k)\text{ that have a }\xi_x<b_k.$$
Wrap Up
Suppose that for some $\epsilon>0$, every $\xi_x>b_k$ for $x\in(a_k,a_k+\epsilon)$. But then $$\max\{f(x+),f(x),f(x-)\}<f(\xi_x)\leq\max\{f(b_k+),f(b_k),f(b_k-)\}$$ for every $x\in(a_k,a_k+\epsilon)$. Letting $\epsilon\to0$ shows that $f(a_k+)\leq \max\{f(b_k+),f(b_k),f(b_k-)\}$.
On the other hand, suppose that for every $\epsilon>0$ there is $x\in(a_k,a_k+\epsilon)$ such that $\xi_x< b_k$. Then (3) above gives $$\max\{f(x+),f(x),f(x-)\}\leq\max\{f(b_k+),f(b_k),f(b_k-)\}$$ Again, taking $\epsilon\to0$ gives $f(a_k+)\leq \max\{f(b_k+),f(b_k),f(b_k-)\}$.

I apologize in advance for any errors I've made. I cannot find any at the moment, but I am happy to correct any that crop up. Anyway, I hope that the spirit of this proof is helpful to you regardless.
