Weighted interval volume

The function $x\mapsto 1\mathrm{/}x$ is positive, finite and smooth on $(0,1)$. On $[a,b]\subset (0,1)$ it is continuous and bounded; the logarithmic integral identity gives its Riemann integral $\log b-\log a=\log (b\mathrm{/}a)$. The Borel Darboux bridge equates the Lebesgue integral to this value. Removing the two null endpoints does not change it, so the chart formula gives ${\mu }_r((a,b))=\log (b\mathrm{/}a)$. For example $(a,b)=(1\mathrm{/}4,3\mathrm{/}4)$ gives $\log 3$.

Every compact subset of $(0,1)$ has finite measure by the positive smooth density theorem. More explicitly, it lies in $[a,b]\subset (0,1)$ and is bounded in measure by $\log (b\mathrm{/}a)$. The increasing sets $K_N=[1\mathrm{/}N,1-1\mathrm{/}N]$, $N\ge 3$, exhaust $(0,1)$ and have mass $\log (N-1)$. Monotone convergence applied to their indicators gives ${\mu }_r((0,1))={\lim }_N\log (N-1)=\mathrm{\infty }$. This divergence also follows without any limit identity for log: each interval $[2^j,2^{j+1}]$ contributes at least $1\mathrm{/}2$ to ${\int }_1^{2^m}dt\mathrm{/}t$, so these logarithms are unbounded.

Empty intervals and singletons have zero mass by the null-set formula. There is no endpoint value of the density at zero or one because neither belongs to the manifold; its blowup at the omitted zero endpoint is consistent with local finiteness.