Interchanging two integrals with different measures. If $\alpha:\mathbb{R} \rightarrow \mathbb{R}$ is a bounded, increasing, right continuous function, and we let $d\alpha$ denote the unique Lebesgue-Stieltjes measure induced by $\alpha$, then I'm trying to show that for any $c \in [0,\infty)$ that
$$\int_{\mathbb{R}}\alpha(x+c)-\alpha(x)dx = c \int_{\mathbb{R}}d\alpha$$
where $dx$ is the Lebesgue integral.
Now just playing around with things and not worrying too much about details, I discovered that
$$\int_{\mathbb{R}}\alpha(x+c)-\alpha(x)dx = \int_{\mathbb{R}}\left(\int_{x}^{x+c}d\alpha\right)dx \stackrel{?}{=} \int_{x}^{x+c}\left(\int_{\mathbb{R}}d\alpha \right)dx = c\int_{\mathbb{R}}d\alpha. $$
My question, is the interchanging of the two integrals (done at the step marked with $\stackrel{?}{=}$) justified? Does Fubini's theorem apply here?
 A: Here is a direct approach using the monotone convergence theorem.
Note that $\alpha$ has limits at $\pm \infty$ (denote by $\alpha(\pm \infty)$ respectively) and we have $\int d \alpha = \alpha(\infty) - \alpha(-\infty)$.
Since $x \mapsto \alpha(x+c) -\alpha(x)$ is non negative it has an integral (possibly infinite).
Let $I_k = \int_{kc}^{(k+1)c} \alpha(x) dx$, then
$\int_{kc}^{(k+1)c} (\alpha(x+c)-\alpha(x)) dx = I_{k+1}-I_k \ge 0$ and
$\int (\alpha(x+c)-\alpha(x)) dx = \sum_k (I_{k+1}-I_k) $.
We see that $\lim_{k \to \infty}I_k = c \alpha(\infty)$,
$\lim_{k \to -\infty}I_k = c \alpha(-\infty)$ and so
$\sum_k (I_{k+1}-I_k) = \lim_{N \to \infty} (I_{N}-I_{-N}) = c (\alpha(\infty)- \alpha(-\infty)) = c \int \alpha$.
A: $\displaystyle\int_{\mathbb{R}}\alpha(x+c)-\alpha(x)dx=\int_{\mathbb{R}}\left(\int_{x}^{x+c}d\alpha\right)dx=\int_{\mathbb{R}}\left(\int_{\mathbb R}\chi_{[x,x+c]}(y) d\alpha(y)\right)dx$
and Tonelli's Theorem implies that
$\displaystyle \int_{\mathbb{R}}\left(\int_{\mathbb R}\chi_{[x,x+c]}(y) d\alpha(y)\right)dx=\int_{\mathbb{R}}\left(\int_{\mathbb R}\chi_{[x,x+c]}(y)dx \right)d\alpha(y)=\int_{\mathbb R}cd\alpha(y)=c\int_{\mathbb R}d\alpha(y)$
