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For a positive random variable $X$, write its base-ten logarithm as an integer part plus a fractional part:The floor function handles values below one as well: for $X=0.0042$, $K=-3$ and the decimal significand is $10^Z=4.2$. Since $10^Z\in [1,10)$, the first nonzero decimal digit is $D=\lfloor 10^Z\rfloor$. The monotonicity of the logarithm gives the exact event equivalenceIf $Z$ has the uniform distribution on $[0,1)$, the probability of this interval is its length. Consequently $\mathbb{P}(D=d)={\log }_{10}(d+1)-{\log }_{10}d$, proving the Benford law. This argument assumes uniformity of the fractional part of the logarithm, rather than uniformity of $X$ itself. See the logarithmic-coordinate explanation in Peter Sheridan Dodds, Benford's Law, essential story.

An exact constructed example is the log-uniform distribution on $[1,10)$, with probability density function $f_X(x)=1/(x\ln 10)$ there and zero elsewhere. Its total mass is ${\int }_1^{10}dx/(x\ln 10)=1$. The change of variables formula with $x=10^z$ gives $f_Z(z)=f_X(10^z)(\ln 10)10^z=1$ for $0\le z<1$. Equivalently, integrating directly givesFor example, the probability of first digit four is ${\log }_{10}(5/4)\approx 0.096910013$. By contrast, an ordinary uniform distribution for $X$ on $[1,10)$ has probability density function $1/9$ there, so $\mathbb{P}(D=d)={\int }_d^{d+1}dx/9=1/9$ for every digit. This is a concrete counterexample to any claim that the Benford law holds for every positive random variable.

More generally, a log-uniform distribution on $[A,B)$ with $0<A<B$ has uniform $Y={\log }_{10}X$ on $[{\log }_{10}A,{\log }_{10}B)$. If its logarithmic span ${\log }_{10}(B/A)=m$ is a positive integer, reduction to the fractional part wraps an interval of length $m$ around the unit interval exactly $m$ times, making $Z$ uniform even when $A$ is not a power of ten. Arbitrary partial-decade truncations generally do not have this property: for a log-uniform distribution on $[1,2)$, the first digit is always one.

Under the uniform fractional-logarithm assumption, multiplication by any constant $c>0$ replaces $Z$ by $Z^{\prime }=\{Z+{\log }_{10}c\}$. This is a rotation modulo one, which preserves the uniform distribution and therefore preserves the Benford law; see scale invariance of decimal significands. For the constructed $[1,10)$ example, multiplying by $0.001$ gives a log-uniform distribution on $[0.001,0.01)$ with the same digit probabilities. The claim concerns the decimal significand law: there is no proper probability density function proportional to $1/x$ over all of $(0,\infty )$, since its integral diverges. The existing scale-invariant prior describes that unnormalizable measure, while a finite log-uniform distribution is normalizable.

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.