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The leading decimal digit is determined by the fractional part of ${\log }_{10}X$. A uniform fractional logarithm gives Benford law by the lengths of the intervals $[{\log }_{10}i,{\log }_{10}(i+1))$. A log-uniform distribution whose logarithmic span is an integer has this property. Arbitrary partial-decade truncations generally do not.

The normalized log-uniform distribution here has density $1/(x\log 10)$. A leading digit $i$ corresponds to $i\le X<i+1$, soThese are exactly the Benford law probabilities. Continuous densities make endpoint conventions immaterial.

The claim needs a complete number of logarithmic decades. For the normalized log-uniform distribution on the specified range, $U={\log }_{10}X$ is uniform on $(a,b)$. Digit $i$ corresponds to $\{U\}\in [{\log }_{10}i,{\log }_{10}(i+1))$. If $b-a=m$ is a positive integer, the integral of a period-one indicator over $(a,a+m)$ is $m$ times its integral over one period. Thus Benford law from logarithmic uniformity givesThis proves the intended case of integers $a<b$, and even permits noninteger $a$ when the span is an integer.

For arbitrary real $a<b$, the exact formula instead isOnly finitely many terms are nonzero. For a counterexample take $a=0$, $b={\log }_{10}2$: then $1<X<2$, so the leading digit is always one. That is not Benford law. The unrestricted range in the PDF needs this qualification.

The measure $d\sigma /\sigma$ is invariant under multiplication by every positive constant. In the logarithmic coordinate it is Lebesgue measure. It cannot be normalized over $(0,\infty )$, although a bounded truncation is a log-uniform distribution.