A property of potential function ($\lim_{z\rightarrow\infty}(p_{\mu}(z)-\mu(\mathbb{C})\log|z|)=O(1/|z|)$)

Question

I am studying the potential theory from Ransford book.

Let $\mu$ be a finite Borel measure on $\mathbb{C}$ with compact support and let $p_{\mu}:\mathbb{C}\rightarrow [-\infty,\infty)$. The result is Theorem 3.1.2 says that

$$p_{\mu}(z)=\int\log|z-w|d\mu(w)=\mu(\mathbb{C})\log|z|+O(1/|z|)\qquad \text{as}\,\,\,z\rightarrow\infty$$

I think I see why the first term of the sum is $\mu(\mathbb{C})\log|z|$. This is because

$$p_{\mu}(z)=\int\log\left|z\cdot\frac{z-w}{z}\right|d\mu(w)=\int\log|z|d\mu(w)+\int\log|1-w/z|d\mu(w)=\mu(\mathbb{C})\log|z|+\color{red}{O(1/|z|)}$$ as $z\rightarrow \infty$

The punch line in the book says that "as $\mu$ has compact support, the final term is $O(1/|z|) as |z|\rightarrow \infty$."

I don't get why the second term in the sum asymptomatically to the Big-Oh of $1/|z|$. It seems easy, but I'm struggling to verify it. Could someone point me out, please?

Answer 1

Let $R>0$ be such that $\text{supp}(\mu)\subset B_R(0)$. Let $|z|>R$. Note that by the triangle inequality $$ 1-R/|z|\leq 1-|w/z|\leq |1-w/z|\leq 1+|w/z|\leq 1+R/|z| $$ Now use the Taylor series for $\log(1\pm x)$ and that our bounds are uniform and the measure is finite.