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

Lemma A: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 $\varphi_n(x) \to 0$.

**Here is the question:** 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 $\xi=1/3$, for instance: \begin{gather} \phi_n(x) = \begin{cases} 1 & x\in \bigl(\tfrac{k}{2^{n}},\tfrac{k+1}{2^{n}}\bigr] \text{ where $k$ is such that } \tfrac{k}{2^{n}}<\tfrac13<\tfrac{k+1}{2^{n}} \\ 0 & \text{elsewhere} \end{cases} \end{gather}

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?

orsome $\phi_k$ is discontinuous. – Pietro Majer Sep 9 '13 at 7:53