Solutions
475
(x, y) ∈ D ⇐⇒ (y, x) ∈ D
c (= E
2
− D)
=⇒ h(y, x) = 1.
Ainsi, apr` es avoir not´ e que
g(x, y) =
(η × Q) 1
(η × Q) 0
(x, y) =
(η × Q) 0
(η × Q) 1
(y, x) =
1
g(y, x)
et
η(dx)g(x, y)Q(x, dy) = g(x, y) × (η × Q) 0 (d(x, y))
= (η × Q) 1 (d(x, y)) = η(dy)Q(y, dx)
on obtient
η(dx)h(x, y)Q(x, dy)f (x)
=
D
η(dx)h(x, y)Q(x, dy)f (x) +
D c
η(dx)h(x, y)Q(x, dy)f (x)
=
D
η(dx)g(x, y)Q(x, dy)f (x) +
D c
η(dx)h(x, y)Q(x, dy)f (x)
=
D
η(dy)Q(y, dx)f (x) +
D c
η(dx)Q(x, dy)f (x)
On montre de mˆ eme que
η(dx)h(x, y)Q(x, dy)f (y)
=
D
η(dx)h(x, y)Q(x, dy)f (y) +
D c
η(dx)h(x, y)Q(x, dy)f (y)
=
D
η(dx)g(x, y)Q(x, dy)f (y) +
D c
η(dx)Q(x, dy)f (y)
=
D
η(dy)Q(y, dx)f (y) +
D c
η(dx)Q(x, dy)f (y)
=
D c
η(dx)Q(x, dy)f (x) +
D
η(dy)Q(y, dx)f (x).
En comparant nos deux r´ esultats, on en conclut que
η(dx)h(x, y)Q(x, dy)f (y) =
η(dx)h(x, y)Q(x, dy)f (x)
Il en d´ ecoule que pour toute fonction born´ ee f , nous avons
(ηM )(f ) = η(f )
et donc la mesure η est bien M -invariante.
8.6.1