# Why is any Lebesgue-Stieltjes measure on $(\mathbb{R},\mathcal{B}_\mathbb{R})$ not complete?

Let $F: \mathbb{R} \to \mathbb{R}$ be any increasing, right-continuous function. Then we define the Lebesgue-Stieltjes measure associated to $F$ to be the unique measure $\mu_F$ on $(\mathbb{R},\mathcal{B}_\mathbb{R})$ such that $\mu_F((a.b])=F(b)-F(a)$ for all $a, b$.

We wish to show that the completion of $\mathcal{B}_\mathbb{R}$ with respect to $\mu_F$, denoted by $M_F$, is strictly larger than $\mathcal{B}_\mathbb{R}$.

Since $\text{card}(\mathcal{B}_\mathbb{R})=\mathfrak{c}$, it will suffice to exhibit a set $K$ in $M_F$ such that $\mu_F(K)=0$ and $\text{card}(K)=\mathfrak{c}$, for then $\text{card}(M_F)>\mathfrak{c}$.

• Do you understand why the usual measure on the Borel algebra of $\mathbb{R}$ is not complete? – Kevin Carlson May 31 '14 at 12:39
• Yes, so I just used pretty much the same argument, thanks! – Aubrey May 31 '14 at 12:42

We mimic the construction of the Cantor set. For any interval $I$, let us call an open subinterval $J$ of $I$ with $\mu_F(J)\geq\mu_F(I)/2$ and endpoints in the interior of $I$ an "open middle superhalf" of $I$. Observe that, since it is increasing, $F$ has only countably many points of discontinuity. Let $K_1$ be a closed interval with endpoints where $F$ is continuous, and $K_{n+1}$ be obtained by removing from each of the intervals comprising $K_n$ an open middle superhalf at whose endpoints $F$ is continuous. Then one can check that $K=\bigcap_{1}^{\infty}K_n$ has the desired properties.