The indicator of a fat Cantor set is upper semicontinuous and equal almost everywhere to no Riemann integrable function

The function ${\chi }_S$ is upper semicontinuous on $[0,1]$. If $x\in S$, then ${\chi }_S(y)\le 1<1+\epsilon$ for every $y$ and every $\epsilon >0$. If $x\mathrm{\notin }S$, then $S$ is closed by [L2], so there is an open neighbourhood of $x$ disjoint from $S$; on that neighbourhood ${\chi }_S$ is identically $0$, which is certainly below $0+\epsilon$. Thus [L1] is satisfied at every point.

Let $g:[0,1]\to \mathbb{R}$ satisfy $g={\chi }_S$ almost everywhere, and let $N$ be a measurable null set containing the disagreement set. Because $S$ is compact, [L3] makes it measurable, and [L2] says it does not have measure zero. Therefore [L4] gives ${\lambda }_1(S\setminus N)={\lambda }_1(S)-{\lambda }_1(S\cap N)={\lambda }_1(S)>0.$ Fix $x\in S\setminus N$. Then $g(x)=1$. Since $S$ is nowhere dense by [L2], every neighbourhood of $x$ meets $[0,1]\setminus S$; the intersection of that neighbourhood with $[0,1]\setminus S$ is a nonempty open set, so it contains a nondegenerate interval and hence has positive measure by [L6]. Because $N$ is null, that positive-measure open set cannot be contained in $N$, so the neighbourhood contains $y\mathrm{\notin }S\cup N$. Then $g(y)={\chi }_S(y)=0$. Thus $g$ is discontinuous at every $x\in S\setminus N$.

Step 2.1 shows that the discontinuity set of $g$ contains the positive-measure set $S\setminus N$. Hence [L5] implies that $g$ is not Riemann integrable. So ${\chi }_S$ is upper semicontinuous but is almost everywhere equal to no Riemann integrable function, and the statement is false.