Exercices : Théorème de convergence - Licence 3

(E, E, µ)λ
(fn)n∈N(E, E, µ)
f
M∈R
∀n∈N,ZE fndµ ≤M
ZE
fdµ ≤M
(fn)n∈N(R,B(R)) (R,B(R)) f2n=
1[0,1
2]f2n+1 =1[1 2,1] µ(R)<+∞
lim sup Zfndµ Zlim sup fndµ
(fn)n∈Nµ
f
lim
n→+∞ZE
|fn−f|dµ = 0 ⇔lim
n→+∞ZE
|fn|dµ =ZE |f|dµ
f g (E, E, µ)
f=g µ Zfdµ =Zgdµ
Zfdµ < +∞, f < +∞µ
Zfdµ = 0, f = 0 µ
∀n∈N,∀x∈[0,1] fn(x) = (1 −x)x2n

Pfnλ[0,1] f
Zfdλ
∞
P
n=0
(−1)n
n+ 1
∀n∈N∗,∀x∈R∗
+fn(x) = e−nx −2e−2nx.
PfnλR∗
+f=
∞
P
n=1
fn.
fnλR∗
+
f
Zfdλ
∞
P
n=1 Zfndλ
f:R→R
lim
n→+∞ZR
e−nsin2(x)f(x)λ(dx)
∀n∈N∗In=Z[1,n] ln 1 + nx
1 + nx3λ(dx)
(In)n
(In)n≥1In=Z[0,n] cosnrx
ne−xλ(dx)
(fn)n∈N∗[0,1]
fn(x) = n2xn(1 −x)
(fn)n∈N∗λ[0,1] f
In=Z[0,1] fndλ n ∈N∗
lim
n→∞ Z[0,1]
fn(x)dλ Z[0,1] lim
n→∞ fn(x)dλ
(fn)n∈N∗sup
n∈N∗
fn
[0; 1]

n In=nZR+ ln 1 + e−x
nλ(dx).
(fn)n≥1[0,1] fn(x) = arctan (nx)
1 + x2
(fn)n≥1λ[0,1] f
Z[0,1]
fn(x)λ(dx)n≥1
∀n∈NIn=ZR+
1 + e−nx
1 + x2λ(dx)
(In)n∈N
x≥0f(x) = ZR+
1
1 + x3+t3λ(dt)
fR+
fR+
fR+
f(t) = ZR+sin(x)
x2
e−txλ(dx)
fR+R∗
+
f00 +∞f
f
f(R,B(R)) µ
f λ
µ
µ
F(x) = µ(] − ∞, x])
F0µ µ λ

α > 0F(x) = (1 −e−αx)1R+(x)
µ
µ
λ(R,B(R))
T:R→RT(x) = 2x+ 5
µTµ T
µTλ
µTµ
Id µT
T(x) = x2