This is a question about the proof of Lemma A in §16 of the book <b>Functional Analysis</b> by F. Riesz and B. Sz.-Nagy.

> <b>Lemma A:</b> For every sequence of step functions $\{\varphi_n\}$ which decreases to zero almost everywhere, the sequence of values of their integrals also tends to zero.

The proof starts, by covering the set $E_0$ of all discontinuities of the sequence, which is of measure zero, with a system of intervals $\Sigma_0$ of total length less than $\varepsilon$. Then it continues:

> For the remaining points $\varphi_n(x) \to 0$.

<b>Here is the question:</b> how do we conclude that the points where the series does not converge are covered by $\Sigma_0$?
It is possible finish the proof by adding the points, where the sequence does not converge to $E_0$, but I assume the authors have a reason not to do so.

On the other hand, I think I have a counterexample: let the discontinuities of the step function $\varphi_n$ be at the points $x_{nk} = k2^{-n}$ for $k=1,\dots,2^{-n}-1$. Let the sequence be convergent except at the point $\xi=1/3$. Choose for $\Sigma_0$ the intervals $I_{nk} = (x_{nk}-\delta_{nk}, x_{nk}+\delta_{nk})$, where
\begin{gather}
\delta_{nk} = \min\left\{\varepsilon2^{-n}2^{-k}, \left|x_{nk}-\tfrac13\right|\right\}.
\end{gather}
Since for all indices $\delta_{nk}>0$, this set covers all discontinuities of the sequence, has total length less than $\varepsilon$, and does not cover the point $\xi$. Or am I wrong here?