Confused about substitution in Stiltjes integral

Question

Suppose we have an integral $$ \int_{-a}^{a} \sin (x) \nu(dx), $$

where $\nu$ is a finite measure with $\nu(-A)=\nu(A), A \in \sigma(\mathbb{R})$ and $x>0$. Then we have $$ \int_{-a}^{a} \sin (x) \nu(dx) = \int_{-a}^0 \sin (x) \nu(dx) + \int_0^{a} \sin (x) \nu(dx). $$

In the first integral we make a substitution $y=-x$, then $$ \int_{-a}^0 \sin (x) \nu(dx) = -\int_{a}^0 \sin (y) \nu(dy) = \int_0^{a} \sin (y) \nu(dy) $$ and $$ \int_{-a}^{a} \sin (x) \nu(dx) = 2 \int_{0}^a \sin (x) \nu(dx), $$ which of course cannot be true. Obviously I have a mistake in the reasoning. Could you please explain me where?

Answer 1

Okay, the whole integration was wrong. The right way to do it is to use, for example, theorem 7 in Chapter II of Shiryaev's Provability. Then $$ \int_{-a}^0 \sin(x) \nu(dx) = \int_0^a \sin(-y) \nu(dy) = -\int_0^a \sin(y) \nu(dy). $$

Answer 2

In yours

$$ \int_{-a}^0 \sin (x) \nu(dx) = -\int_{a}^0 \sin (y) \nu(dy) $$

there is a mistake: both $\sin$ and $\nu$ shall give you the minus, so the minus does not appear here.