Natural density implies Dirichlet density

Apply [L2] to the coefficients $a_n={\mathbf{1}}_A(n)$. Then for $s>1$ one has ${\sum }_{n\in A}{n}^{-s}=s{\int }_1^{\mathrm{\infty }}A(x)x^{-s-1}dx$. Now write $A(x)=\delta x+E(x)$ with $E(x)=o(x)$ by [L1].

Substituting into step 1.1 gives ${\sum }_{n\in A}{n}^{-s}=\delta s{\int }_1^{\mathrm{\infty }}{x}^{-s}dx+s{\int }_1^{\mathrm{\infty }}E(x)x^{-s-1}dx=\delta s\mathrm{/}(s-1)+s{\int }_1^{\mathrm{\infty }}E(x)x^{-s-1}dx$. It therefore suffices to show that $(s-1){\int }_1^{\mathrm{\infty }}E(x)x^{-s-1}dx\to 0$ as $s\downarrow 1$.

Fix $\epsilon >0$. Because $E(x)=o(x)$, choose $X\ge 1$ so that $\mathrm{\mid }E(x)\mathrm{\mid }\le \epsilon x$ for every $x\ge X$. For $1<s\le 2$, the quantity $\mid (s-1)s{\int }_1^{\mathrm{\infty }}E(x)x^{-s-1}dx\mid$ is at most $(s-1)s{\int }_1^X\mathrm{\mid }E(x)\mathrm{\mid }{x}^{-s-1}dx+\epsilon s(s-1){\int }_X^{\mathrm{\infty }}{x}^{-s}dx$. The first term tends to $0$ because the integral over $[1,X]$ is bounded, and the second term is at most $2\epsilon X^{1-s}\le 2\epsilon$. Since $\epsilon$ is arbitrary, the whole expression tends to $0$. Hence $(s-1){\sum }_{n\in A}{n}^{-s}\to \delta$, which is exactly the Dirichlet-density statement from Natural and Dirichlet density.