TD5 : 08/02/2017 Correction Exos Théorie de la mesure

ε > 0λ2(B(0, ε)) >0B(0; ε)
R2εR2
λ2([a, b]×[c, d]) = (b−a)×(b−c).
ε > 0 0 < δ < ε
√2([−δ, δ]×[−δ, δ]) ⊂B(0, ε)
λ2
λ2([−δ, δ]×[−δ, δ]) 6λ2(B(0, ε)) ⇔(2δ)×(2δ)6λ2(B(0, ε))
λ2(B(0, ε)) >4δ2>0∀ε > 0
λ2(H1={(x, y)∈R2, x = 1}) = 0?
H1n
H1=∪n∈NHn
1, Hn 1=(x, y)∈R2x= 1 y∈[−n, n]
Hn
1ε
Hn
1=∩ε>0An εAn ε=(x, y)∈R2x∈]1 −ε, 1 + ε]y∈[−n, n].
λ2(An
ε) = (2n)×(2ε)=4nε ε 7→ 0
λ2(Hn
1) = λ2(∩ε>0An ε) = lim
ε7→0λ2(An
ε) = lim
ε7→04nε = 0.
n7→ +∞
λ2(H1) = lim
n7→+∞
λ2(Hn
1)=0

λ2(D={(x, y)∈R2|x=y}) = 0?
D n
D=∪n∈NDn, Dn=(x, y)∈R2|x=y, x ∈[−n, n].
DnDn⊆Pn
k
Pn
k=∪i=k−1
i=−kni
k,n(i+ 1) k×ni k,n(i+ 1) k
λ2
λ2(Dn)6λ2(Pn
k)
Pn
k
λ2(Pn
k) =
i=k−1
X
i=−kn(i+ 1)
k−ni k.n(i+ 1) k−ni k=
i=k−1
X
i=−k
n
k.n k
=
k−1
X
i=−k
n2
k2= ((k−1) + k+ 1) .n2 k2= 2k.n2 k2=2n2 k.
λ2(Dn)6lim
k7→+∞
λ2(Pn
k) = 0
λ2(Dn) = 0 λ2(D) = lim
n7→+∞
λ2(Dn)=0
fR2;ZR f(x, x)dx < +∞.
f(x, y) = (0B(0,1)
1
√2πy2e
−(x−y)2
2y8Bc(0,1).
f f(x, y)>0∀(x, y)∈R2
fR2
ZR2
f(x, y)dx dy =ZB(0,1) f(x, y)dxdy +ZBc(0,1) f(x, y)dxdy = 0 + ZBc(0,1) f(x, y)dxdy
=ZBc(0,1)
1
√2πy2e
−(x−y)2
2y8dxdy
=Zy2Z1
√2πe
−(x−y)2
2y8dxdy ∀(x, y)∈Bc(0,1)
=Zy2Z1
√2πe−1
2(x−y
y
−4)2dxdy ∀(x, y)∈Bc(0,1)
=Zy2 Zy−4
y−4
e−1
2(x−y
y
−4)2
√2πdx!dy ∀(x, y)∈Bc(0,1)
=Zy−2Z1
√2πy−4e−1
2(x−y
y
−4)2dxdy ∀(x, y)∈Bc(0,1)

N(m, σ)
N(m, σ)
f(x) = 1
√2πσ e
−(x−m)2
2σ2
σ=y−4m=y
Z]−∞,−1[×]1,+∞[
1
√2πy−4e−1
2(x−y
y
−4)2dx = 1 −α < ∞, α > 0.
ZR2
f(x, y)dx dy =Zy−2Z1
√2πy−4e−1
2(x−y
y
−4)2dxdy ∀(x, y)∈Bc(0,1)
= (1 −α)Z]−∞,−1[×]1,+∞[ y−2dy < ∞
x=y
f(x, x) = (0B(0,1)
1
√2πx2e
−(x−x)2
2x8Bc(0,1).=0B(0,1)
1
√2πx2Bc(0,1).
ZR
f(x, x)dx = +∞.