$\int_{-1}^0 f d\alpha, \int_0^1 f d\alpha$ both exists, but not $\int_{-1}^{1} f d\alpha$ (Stieltjes) 
Let $f,\alpha:[a,b]\to\mathbb{R}$ be bounded, such that $\int_a^b f
 d\alpha$ exists. Then, for each $c\in [a,b]$, the integrals $\int_a^c
 f d\alpha$ and $\int_a^b f d\alpha$ exists and $\int_a^b f d\alpha =
 \int_a^c f d\alpha + \int_c^b f d\alpha$. Consider $f,\alpha$:[-1,1]
  defined by $f(x) = 0$ if $-1\le x < 0$, $f(x) = 1$ if $0\le x \le 1$,
  $\alpha(x) = 0$ if $-1\le x \le 0$ and $\alpha(x) = 1$ if $0< x \le
 1$. Show that both exist:
$$\int_{-1}^0 f d\alpha, \int_0^1 f d\alpha$$
but
$$\int_{-1}^{1} f d\alpha$$
does not exist.

I don't understand this question. It seems that when $f,\alpha:[a,b]\to\mathbb{R}$ are both bounded and the integral $\int f d\alpha$ exists, then $\int_a^b f d\alpha =
 \int_a^c f d\alpha + \int_c^b f d\alpha$. Then, it gives me an example of $f$ and $\alpha$ both bounded and ask me to prove that the relation above isn't valid. But this would be a contradiction... What's happening?
 A: $$
\int_{-1}^{0}f\,d\alpha=0
$$
Proof: Let $-1=t_{0}<t_{1}<\ldots<t_{n}=0$ be an arbitrary partition
of $[-1,0]$ and let $\xi_{i}\in[t_{i-1},t_{i}]$ be arbitrary, $i=1,\ldots,n$.
Consider the Riemann-Stieltjes sum: 
$$
\sum_{i=1}^{n}f(\xi_{i})\left(\alpha(t_{i})-\alpha(t_{i-1})\right).
$$
By the definition of $\alpha$, we always have $\alpha(t_{i})-\alpha(t_{i-1})=0$,
hence the sum is zero. It follows that the Riemann-Stieltjes integral
is zero.
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$$
\int_{0}^{1}f\,d\alpha=1
$$
Proof: Let $0=t_{0}<t_{1}<\ldots<t_{n}=1$ be a partition of $[0,1]$.
Let $\xi_{i}\in[t_{i-1},t_{i}]$. Then 
$$
\sum_{i=1}^{n}f(\xi_{i})\left(\alpha(t_{i})-\alpha(t_{i-1})\right)=1
$$
because $f(\xi_{i})=1$, $\alpha(t_{1})-\alpha(t_{0})=1$ and $\alpha(t_{i})-\alpha(t_{i-1})=0$
for $i>1$. Therefore the integral exists and is equal to $1$.
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The integral 
$$
\int_{-1}^{1}f\,d\alpha
$$
does not exist.
Proof: Prove by contradiction. Suppose that the integral exists and
equal to $I$. For $\varepsilon=0.01$, there exists $\delta>0$ such
that for any partition $-1=t_{0}<t_{1}<\ldots<t_{n}=1$ and any $\xi_{i}\in[t_{i-1},t_{i}]$,
if 
$$
\max_{1\leq i\leq n}|t_{i}-t_{i-1}|<\delta,
$$
then 
$$
|\sum_{i=1}^{n}f(\xi_{i})\left(\alpha(t_{i})-\alpha(t_{i-1})\right)-I|<\epsilon.
$$
Choose a partition $-1=t_{0}<t_{1}<\ldots<t_{n}=1$ such that $\max_{1\leq i\leq n}|t_{i}-t_{i-1}|<\delta$
and $0\in(t_{k-1},t_{k})$ for some $k$. Clearly such a partition exists. Construct two sums by choosing
$\xi_{i}\in[t_{i-1},t_{i}]$ in two different way. For $i\neq k$,
choose $\xi_{i}$ in an arbitrary way. For the first sum, we choose
$\xi_{k}\in(t_{k-1},0)$. For the second sum, we choose $\xi_{k}\in(0,t_{k})$.
Note that for these two sums, we always have 
$$
\sum_{i=1}^{n}f(\xi_{i})\left(\alpha(t_{i})-\alpha(t_{i-1})\right)=f(\xi_{k})\left(\alpha(t_{k})-\alpha(t_{k-1})\right)=f(\xi_{k})
$$
because for $i\neq k$, $\alpha(t_{i})-\alpha(t_{i-1})=0$ and $\alpha(t_k)-\alpha(t_{k-1})=1$. Now it
is clear that the first sum equals to $0$ while the second sum equals
to $1$. Hence we have $|0-I|<0.01$ and $|1-I|<0.01$ which is
a contradiction.
Therefore 
$$
\int_{-1}^{1}f\,d\alpha
$$
does not exist.
