I am reading "Analysis on Manifolds" by James R. Munkres.
There is the following exercise on p.111 in this book.
Exercise 6.
Show that Theorem 13.6 holds without the hypothesis that $f$ is continuous on $S$.
My solution is here:
Since Munkres doesn't use the hypothesis that $f$ is continuous on $S$ in his proof, Theorem 13.6 holds without the hypothesis.
My solution is too simple, so I don't have perfect confidence in my solution.
I think Munkres doesn't use the hypothesis that $f$ is continuous on $S$ in his proof at all.
Am I right or not?
Best Answer
To prove that $f_A = f \chi_A$ is Riemann integrable, it is sufficient that $f_A$ be bounded and continuous almost everywhere in a rectangle $Q \supset S \supset A$. The hypothesis that $f_S$ is Riemann integrable is enough since it means that $f_S$ is bounded and continuous almost everywhere in $Q$.
Because $|f_A| \leqslant |f_S|$, we immediately have that $f_A$ is bounded.
The argument in Step 1 then proves that $f_A$ is continuous everywhere in $Q$ except possibly on a set of measure zero. Again this follows from the assumption that $f_S$ is Riemann integrable without a stronger assumption that $f$ is continuous.
This is obvious for $x_0 \in A = \mathrm{Int }\, S$ and $x_0 \in \mathrm{Ext }\,\bar{S}$ since there exist neighborhoods of $x_0$ where $f_A = f_S$, and either $f_S$ is continuous at $x_0$ or $x_0$ belongs to a set of measure zero. The argument is more involved but striaghtforward when $x_0$ is a boundary point. Since $f_A(x_0) = 0$, there is a discontinuity at $x_0$ only if $f_A(x) = f_S(x) \not \to 0$ as $x \to x_0$ from the interior, that is with $x \in A$. However, such points are discontinuity points for $f_S$ and must be contiened in a set of measure zero as well.