Contrôle continu 2 Corrigé Intégrale et théorêmes de

(RXfndµ)n∈Nsup
5
R+∞
−∞ sin( x
n)n x(1+x2)dx
sin( x
n)n x1x→0n
R|sin( x
n)| ≤ | x n| | sin( x n)n x(1+x2)| ≤ 1 1+x2Rx
fn(x) = sin( x
n)n x(1+x2)→1 1+x2n→+∞fn(x)
n
Z+∞
−∞
fn(x)dx →Z+∞
−∞
1
1 + x2dx = [arctan]+∞
−∞ =π
Pk≥01
3k(1 −1
n(k+1) )fn(x) = 1 3x(1 −1 n(x+1) )µN
X
k≥0
1
3k(1 −1 n(k+ 1)) = ZN
fn(x)dµ
µ1
3k(1 −1
n(k+1) )≤1
3k
n fn(x)→1
3xn→+∞
ZN
fn(x)dµ →X
k≥0
1
3k=3 2
I(α) = lim
n→+∞Zn
0
(1 −x
n)neαxdx.
fn(x) = (1 −x
n)neαx10≤x≤nfn+1(x)−fn(x)x>n+ 1
n≤x≤n+ 1 fn+1(x) 1 −x
n+1 ≥0
0< x < n
fn+1(x)−fn(x) = eαx((1 −x
n+ 1)n+1 −(1 −x n)n) = eαx(e(n+1) ln(1−x
n+1 )−enln(1−x
n))
eαx g[0, n[
g0(x) = −1
1−x
n+1
+1
1−x
n
1−x
n+1 ≥1 1−x
n≥01
1−x
n≥1
1−x
n+1
g[0, n[x→n+∞
x→0 [0, n[g(n+ 1) ln(1 −x
n+1 )≥nln(1 −x n)
e(n+1) ln(1−x
n+1 )≥enln(1−x
n)fn+1(x)−fn(x)≥0 [0, n[
fn
fnfn(x) = enln(1−x
n)eαx10≤x≤n
fn(x)→e−x+αx =e(−1+α)x−1 + α < 0α < 1
α < 1I(α) = 1
1−αI(α) = +∞

f(Rn,B(Rn), µ)τyf y τyf(x) =
f(x−y)
τ(y) = ZRn |τyf−f|dµ →0
y→0f
f f
f=1AA∈ B(Rd)τyf(x) = 1A(x−y) = 1y+A
1A+y(x)−1A(x) = 
0A∩A+y
1A+y\A
−1A\A+y
τ(y) = µ(A+y\A) + µ(A\A+y)
yn0
τ(yn) = µ(A+yn\A) + µ(A\A+yn)
µ(A) lim sup µ(A\A+yn)≤µ(lim sup A\A+yn)=0
yn0B A +yn\A⊂B µ(B)<+∞
lim sup µ(A+yn\A)≤µ(lim sup A+yn\A)=0
τ(yn)→0 (yn)n∈N0f
τ(y)→0y→0
f
f  > 0g g ≤f
ZRn
|f−g| ≤ 
3
ZRn
|τyf−τyg| ≤ 
3
|τ(y)| ≤ ZRn |τyf−τyg|+ZRn |τyg−g|+ZRn |f−g|
g≤f g η > 0y
||y|| < η
ZRn
|τyg−g| ≤ 
3
|τ(y)| ≤ 
f=f+−f−f+=max(f, 0) f−=max(−f, 0) τyf=τyf+−τyf−