(A1) Sejam m e M; menor e maior "altura" do retângulo de base b-a Sejam
m
=
inf
{
f
(
x
)
;
x
∈
[
a
,
b
]
}
e
M
=
sup
{
f
(
x
)
;
x
∈
[
a
,
b
]
}
{\displaystyle m={\mbox{inf}}\{f(x);x\in [a,b]\}{\mbox{ e }}M={\mbox{sup}}\{f(x);x\in [a,b]\}}
m
(
b
−
a
)
≤
M
(
b
−
a
)
{\displaystyle m(b-a)\leq M(b-a)}
P
0
=
{
a
,
b
}
temos
S
_
(
f
;
P
0
)
≤
S
¯
(
f
;
P
0
)
{\displaystyle P_{0}=\{a,b\}{\mbox{ temos }}{\underline {S}}(f;P_{0})\leq {\overline {S}}(f;P_{0})}
(A2) Sejam
m
i
e
M
i
{\displaystyle m_{i}\;e\;M_{i}}
t
i
−
t
i
−
1
{\displaystyle t_{i}-t_{i-1}}
Podemos calcular a área da partição
P
1
{\displaystyle P_{1}}
Por falta
A
(
f
a
l
t
a
)
=
m
1
(
t
1
−
t
0
)
+
m
2
(
t
2
−
t
1
)
=
S
_
(
f
;
P
1
)
{\displaystyle A(falta)=m_{1}(t_{1}-t_{0})+m_{2}(t_{2}-t_{1})={\underline {S}}(f;P_{1})}
Onde
m
1
=
i
n
f
{
f
(
x
)
;
x
∈
[
t
0
,
t
1
]
}
e
m
2
=
i
n
f
{
f
(
x
)
;
x
∈
[
t
1
,
t
2
]
}
{\displaystyle m_{1}=inf\{f(x);x\in [t_{0},t_{1}]\}{\mbox{ e }}m_{2}=inf\{f(x);x\in [t_{1},t_{2}]\}}
Por sobra
A
(
s
o
b
r
a
)
=
M
1
(
t
1
−
t
0
)
+
M
2
(
t
2
−
t
1
)
=
S
¯
(
f
;
P
1
)
{\displaystyle A(sobra)=M_{1}(t_{1}-t_{0})+M_{2}(t_{2}-t_{1})={\overline {S}}(f;P_{1})}
Onde
M
1
=
s
u
p
{
f
(
x
)
;
x
∈
[
t
0
,
t
1
]
}
e
M
2
=
s
u
p
{
f
(
x
)
;
x
∈
[
t
1
,
t
2
]
}
{\displaystyle M_{1}=sup\{f(x);x\in [t_{0},t_{1}]\}{\mbox{ e }}M_{2}=sup\{f(x);x\in [t_{1},t_{2}]\}}
Como
m
1
≤
M
1
e
m
2
≤
M
2
{\displaystyle m_{1}\leq M_{1}{\mbox{ e }}m_{2}\leq M_{2}}
S
_
(
f
;
P
1
)
≤
S
¯
(
f
;
P
1
)
{\displaystyle {\underline {S}}(f;P_{1})\leq {\overline {S}}(f;P_{1})}
(A3) Seja
m
(
b
−
a
)
=
m
(
t
2
−
t
0
)
. Tomando
t
1
∈
[
t
0
,
t
2
]
, temos
m
(
t
2
−
t
1
+
t
1
−
t
0
)
=
{\displaystyle m(b-a)=m(t_{2}-t_{0}){\mbox{. Tomando }}t_{1}\in [t_{0},t_{2}]{\mbox{, temos }}m(t_{2}-t_{1}+t_{1}-t_{0})=}
=
m
(
t
2
−
t
1
)
+
m
(
t
1
−
t
0
)
≤
m
2
(
t
2
−
t
1
)
+
m
1
(
t
1
−
t
0
)
⇒
S
_
(
f
;
P
0
)
≤
S
_
(
f
;
P
1
)
, pois
m
≤
m
1
e
m
≤
m
2
{\displaystyle =m(t_{2}-t_{1})+m(t_{1}-t_{0})\leq m_{2}(t_{2}-t_{1})+m_{1}(t_{1}-t_{0})\Rightarrow {\underline {S}}(f;P_{0})\leq {\underline {S}}(f;P_{1}){\mbox{, pois }}m\leq m_{1}{\mbox{ e }}m\leq m_{2}}
(A4) o fato que
S
¯
(
f
;
P
1
)
≤
S
¯
(
f
;
P
0
)
{\displaystyle {\overline {S}}(f;P_{1})\leq {\overline {S}}(f;P_{0})}
(A2),(A3)e(A4)
S
_
(
f
;
P
0
)
≤
S
_
(
f
;
P
1
)
≤
S
¯
(
f
;
P
1
)
≤
S
¯
(
f
;
P
0
)
{\displaystyle {\underline {S}}(f;P_{0})\leq {\underline {S}}(f;P_{1})\leq {\overline {S}}(f;P_{1})\leq {\overline {S}}(f;P_{0})}
Pelo que vimos acima, quando acrescentamos um único ponto a partição inicial [a,b], a nossa soma inferior ficou maior, e nossa soma superior ficou menor. A nossa idéia então é fazer com que elas se aproximem o suficiente até
|
S
¯
(
f
;
P
∞
)
−
S
_
(
f
;
P
∞
)
|
<
ϵ
,
o
n
d
e
P
∞
{\displaystyle |{\overline {S}}(f;P_{\infty })-{\underline {S}}(f;P_{\infty })|<\epsilon ,onde\;P_{\infty }}
lim
P
i
→
P
∞
|
t
i
−
t
i
−
1
|
=
0
{\displaystyle \lim _{P_{i}\to P_{\infty }}|t_{i}-t_{i-1}|=0}
Relações entre partição e subpartição
Editar
Lema 1 (refinando uma partição)
Editar
Sejam
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\rightarrow \mathbb {R} }
P
k
−
1
, Q cujo
Q
=
P
k
−
1
∪
{
c
}
{\displaystyle P_{k-1}{\mbox{, Q cujo }}Q=P_{k-1}\cup \{c\}}
S
_
(
f
;
P
k
−
1
)
≤
S
_
(
f
;
Q
)
≤
S
¯
(
f
;
Q
)
≤
S
¯
(
f
;
P
k
−
1
)
{\displaystyle {\underline {S}}(f;P_{k-1})\leq {\underline {S}}(f;Q)\leq {\overline {S}}(f;Q)\leq {\overline {S}}(f;P_{k-1})}
Demonstração
Editar
Sejam
P
k
−
1
=
{
t
0
,
t
1
,
.
.
.
,
t
l
−
1
,
t
l
,
t
l
+
1
,
.
.
.
t
k
−
1
,
t
k
}
e
Q
=
P
k
−
1
∪
{
c
}
onde
c
∈
[
t
l
−
1
,
t
l
]
{\displaystyle P_{k-1}=\{t_{0},t_{1},...,t_{l-1},t_{l},t_{l+1},...t_{k-1},t_{k}\}{\mbox{ e }}Q=P_{k-1}\cup \{c\}{\mbox{ onde }}c\in [t_{l-1},t_{l}]}
S
_
(
f
;
P
k
−
1
)
=
∑
i
=
1
k
m
i
(
t
i
−
t
i
−
1
)
=
∑
i
=
1
l
−
1
m
i
(
t
i
−
t
i
−
1
)
+
m
l
(
t
l
−
t
l
−
1
)
+
∑
i
=
l
k
m
i
(
t
i
−
t
i
−
1
)
{\displaystyle {\underline {S}}(f;P_{k-1})=\sum _{i=1}^{k}m_{i}(t_{i}-t_{i-1})=\sum _{i=1}^{l-1}m_{i}(t_{i}-t_{i-1})+m_{l}(t_{l}-t_{l-1})+\sum _{i=l}^{k}m_{i}(t_{i}-t_{i-1})}
S
_
(
f
;
Q
)
=
∑
i
=
1
l
−
1
m
i
(
t
i
−
t
i
−
1
)
+
m
′
(
c
−
t
l
−
1
)
+
m
″
(
t
l
−
c
)
+
∑
i
=
l
k
m
i
(
t
i
−
t
i
−
1
)
{\displaystyle {\underline {S}}(f;Q)=\sum _{i=1}^{l-1}m_{i}(t_{i}-t_{i-1})+m'(c-t_{l-1})+m''(t_{l}-c)+\sum _{i=l}^{k}m_{i}(t_{i}-t_{i-1})}
Onde
m
′
=
i
n
f
{
f
(
x
)
;
x
∈
[
t
l
−
1
,
c
]
}
e
m
″
=
i
n
f
{
f
(
x
)
;
x
∈
[
c
,
t
i
]
}
{\displaystyle m'=inf\{f(x);x\in [t_{l-1},c]\}{\mbox{ e }}m''=inf\{f(x);x\in [c,t_{i}]\}}
É verdade que
m
l
(
t
l
−
t
l
−
1
)
=
m
l
(
t
l
−
c
+
c
−
t
l
−
1
)
=
m
l
(
t
l
−
c
)
+
m
l
(
c
−
t
l
−
1
)
como
m
l
≤
m
′
e
m
l
≤
m
″
{\displaystyle m_{l}(t_{l}-t_{l-1})=m_{l}(t_{l}-c+c-t_{l-1})=m_{l}(t_{l}-c)+m_{l}(c-t_{l-1}){\mbox{ como }}m_{l}\leq m'{\mbox{ e }}m_{l}\leq m''}
S
_
(
f
;
P
k
−
1
)
≤
S
_
(
f
;
Q
)
{\displaystyle {\underline {S}}(f;P_{k-1})\leq {\underline {S}}(f;Q)}
De forma análoga se demonstra que
S
¯
(
f
;
Q
)
≤
S
¯
(
f
;
P
k
−
1
)
{\displaystyle {\overline {S}}(f;Q)\leq {\overline {S}}(f;P_{k-1})}
Sejam
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\rightarrow \mathbb {R} }
Demonstração
Editar
Pelo Lema 1, Q é uma refinação da partição P.
Sejam
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\rightarrow \mathbb {R} }
S
_
(
f
;
P
)
≤
S
¯
(
f
;
Q
)
{\displaystyle {\underline {S}}(f;P)\leq {\overline {S}}(f;Q)}
Demonstração
Editar
Refinando P nos pontos de Q, e refinando Q nos pontos de P teremos
S
_
(
f
;
P
)
≤
S
_
(
f
;
P
∪
Q
)
e
S
¯
(
f
;
P
∪
Q
)
≤
S
¯
(
f
;
P
)
{\displaystyle {\underline {S}}(f;P)\leq {\underline {S}}(f;P\cup Q){\mbox{ e }}{\overline {S}}(f;P\cup Q)\leq {\overline {S}}(f;P)}
S
_
(
f
;
P
∪
Q
)
≤
S
¯
(
f
;
P
∪
Q
)
⇒
S
_
(
f
;
P
)
≤
S
_
(
f
;
P
∪
Q
)
≤
S
¯
(
f
;
P
∪
Q
)
≤
S
¯
(
f
;
P
)
{\displaystyle {\underline {S}}(f;P\cup Q)\leq {\overline {S}}(f;P\cup Q)\Rightarrow {\underline {S}}(f;P)\leq {\underline {S}}(f;P\cup Q)\leq {\overline {S}}(f;P\cup Q)\leq {\overline {S}}(f;P)}
Seja
f
:
[
a
,
b
]
→
R
limitada e
P
∗
{\displaystyle f:[a,b]\rightarrow \mathbb {R} {\mbox{ limitada e }}P^{*}}
∫
_
a
b
f
(
x
)
d
x
=
sup
S
_
(
f
;
P
∗
)
{\displaystyle {\underline {\int }}_{a}^{b}f(x)\,dx={\mbox{ sup }}{\underline {S}}(f;P^{*})}
∫
¯
a
b
f
(
x
)
d
x
=
inf
S
¯
(
f
;
P
∗
)
{\displaystyle {\overline {\int }}_{a}^{b}f(x)\,dx={\mbox{ inf }}{\overline {S}}(f;P^{*})}
Pelo Lema 1
S
_
(
f
;
P
∗
)
≤
sup
S
_
(
f
;
P
∗
)
≤
inf
S
¯
(
f
;
P
∗
)
≤
S
¯
(
f
;
P
∗
)
{\displaystyle {\underline {S}}(f;P^{*})\leq {\mbox{ sup }}{\underline {S}}(f;P^{*})\leq {\mbox{ inf }}{\overline {S}}(f;P^{*})\leq {\overline {S}}(f;P^{*})}
Logo
S
_
(
f
;
P
∗
)
≤
∫
_
a
b
f
(
x
)
d
x
≤
∫
¯
a
b
f
(
x
)
d
x
≤
S
¯
(
f
;
P
∗
)
{\displaystyle {\underline {S}}(f;P^{*})\leq {\underline {\int }}_{a}^{b}f(x)dx\leq {\overline {\int }}_{a}^{b}f(x)dx\leq {\overline {S}}(f;P^{*})}
Lema 2 (soma conservada no refinamento)
Editar
Seja
c
∈
]
a
,
b
[
{\displaystyle c\in ]a,b[}
Q
∗
{\displaystyle Q^{*}}
Q
∗
=
P
∗
∪
{
c
}
{\displaystyle Q^{*}=P^{*}\cup \{c\}}
∫
_
a
b
f
(
x
)
d
x
,
∫
¯
a
b
f
(
x
)
d
x
{\displaystyle {\underline {\int }}_{a}^{b}f(x)dx,{\overline {\int }}_{a}^{b}f(x)dx}
Demonstração
Editar
Em particular
Q
⊂
Q
∗
{\displaystyle Q\subset Q^{*}}
Seja
P
=
Q
∖
c
{\displaystyle P=Q\setminus {c}}
P
⊂
P
∗
{\displaystyle P\subset P^{*}}
Pelo Lema 1
S
_
(
f
;
P
)
≤
S
_
(
f
;
Q
)
≤
S
¯
(
f
;
Q
)
≤
S
¯
(
f
;
P
)
{\displaystyle {\underline {S}}(f;P)\leq {\underline {S}}(f;Q)\leq {\overline {S}}(f;Q)\leq {\overline {S}}(f;P)}
olhemos para o fato que A' = {cota inferior de Q} e B' = {cota superior de Q}; A = {cota inferior de P} e B = {cota superior de P}
A
⊂
A
′
⇒
s
e
a
∈
A
e
a
′
∈
A
′
,
l
o
g
o
a
≤
a
′
{\displaystyle A\subset A'\Rightarrow se\;a\in A\;e\;a'\in A',logo\;a\leq a'}
sup A = sup A', pois
c
∈
]
a
,
b
[
{\displaystyle c\in ]a,b[}
B
⊂
B
′
⇒
s
e
b
∈
B
e
b
′
∈
B
′
,
l
o
g
o
b
≤
b
′
{\displaystyle B\subset B'\Rightarrow se\;b\in B\;e\;b'\in B',logo\;b\leq b'}
inf B = inf B', pois
c
∈
]
a
,
b
[
{\displaystyle c\in ]a,b[}
sup
S
_
(
f
;
P
)
=
sup
S
_
(
f
;
Q
)
≤
inf
S
¯
(
f
;
Q
)
=
inf
S
¯
(
f
;
P
)
{\displaystyle {\mbox{ sup }}{\underline {S}}(f;P)={\mbox{ sup }}{\underline {S}}(f;Q)\leq {\mbox{ inf }}{\overline {S}}(f;Q)={\mbox{ inf }}{\overline {S}}(f;P)}
Sejam A, B subconjuntos não vazios e limitados dos reais. (a) => (b)
(a) Se
A
=
{
a
∈
A
}
,
B
=
{
b
∈
B
}
{\displaystyle A=\{a\in A\},B=\{b\in B\}}
A
+
B
=
{
a
+
b
;
a
∈
A
,
b
∈
B
}
{\displaystyle A+B=\{a+b;a\in A,b\in B\}}
(b) inf(A+B) = inf A + inf B ; sup(A+B) = sup A + Sup B Demonstração
Editar
Dado
a
∈
A
,
b
∈
B
,
t
e
m
o
s
a
≥
i
n
f
A
,
b
≥
i
n
f
B
⇒
a
+
b
≥
i
n
f
A
+
i
n
f
B
{\displaystyle a\in A,b\in B,temos\;a\geq infA,b\geq infB\Rightarrow a+b\geq infA+infB}
Assim inf A + inf B é uma cota inferior de A+B, Dado
ϵ
>
0
,
∃
a
′
∈
A
,
b
′
∈
B
;
a
′
<
i
n
f
A
+
ϵ
2
,
b
′
<
i
n
f
B
+
ϵ
2
⇒
a
′
+
b
′
<
i
n
f
A
+
i
n
f
B
+
ϵ
{\displaystyle \epsilon >0,\exists a'\in A,b'\in B;a'<infA+{\epsilon \over 2},b'<infB+{\epsilon \over 2}\Rightarrow a'+b'<infA+infB+\epsilon }
portanto inf A + inf B é o ínfimo do conjunto A + B o sup se mostra analogamente Sejam
f
,
g
:
[
a
,
b
]
↦
R
{\displaystyle f,g:[a,b]\mapsto \mathbb {R} }
(a)
s
u
p
(
f
+
g
)
≤
s
u
p
(
f
)
+
s
u
p
(
g
)
{\displaystyle sup(f+g)\leq sup(f)+sup(g)}
(b)
i
n
f
(
f
+
g
)
≥
i
n
f
(
f
)
+
i
n
f
(
g
)
{\displaystyle inf(f+g)\geq inf(f)+inf(g)}
Demonstração
Editar
Se
A
=
f
(
[
a
,
b
]
)
,
B
=
g
(
[
a
,
b
]
)
{\displaystyle A=f([a,b]),\;B=g([a,b])}
C
=
{
f
(
x
)
+
g
(
x
)
;
x
∈
[
a
,
b
]
}
⊂
A
+
B
{\displaystyle C=\{f(x)+g(x);x\in [a,b]\}\subset A+B}
pelo teorema
i
n
f
(
A
+
B
)
≤
i
n
f
(
C
)
≤
s
u
p
(
C
)
≤
s
u
p
(
A
+
B
)
{\displaystyle inf(A+B)\leq inf(C)\leq sup(C)\leq sup(A+B)}
lema 3 temos
(a)
s
u
p
(
f
+
g
)
=
s
u
p
(
C
)
≤
s
u
p
(
A
+
B
)
=
s
u
p
(
A
)
+
s
u
p
(
B
)
=
s
u
p
(
f
)
+
s
u
p
(
g
)
{\displaystyle sup(f+g)=sup(C)\leq sup(A+B)=sup(A)+sup(B)=sup(f)+sup(g)}
(b)
i
n
f
(
f
+
g
)
=
i
n
f
(
C
)
≥
i
n
f
(
A
+
B
)
=
i
n
f
(
A
)
+
i
n
f
(
B
)
=
i
n
f
(
f
)
+
i
n
f
(
g
)
{\displaystyle inf(f+g)=inf(C)\geq inf(A+B)=inf(A)+inf(B)=inf(f)+inf(g)}
Sejam
c
∈
[
a
,
b
]
,
f
:
[
a
,
b
]
↦
R
{\displaystyle c\in [a,b],f:[a,b]\mapsto \mathbb {R} }
(a)
∫
_
a
b
f
(
x
)
d
x
=
∫
_
a
c
f
(
x
)
d
x
+
∫
_
c
b
f
(
x
)
d
x
{\displaystyle {\underline {\int }}_{a}^{b}f(x)dx={\underline {\int }}_{a}^{c}f(x)dx+{\underline {\int }}_{c}^{b}f(x)dx}
(b)
∫
¯
a
b
f
(
x
)
d
x
=
∫
¯
a
c
f
(
x
)
d
x
+
∫
¯
c
b
f
(
x
)
d
x
{\displaystyle {\overline {\int }}_{a}^{b}f(x)dx={\overline {\int }}_{a}^{c}f(x)dx+{\overline {\int }}_{c}^{b}f(x)dx}
Demonstração
Editar
(a)Sejam
A
=
{
S
_
(
f
/
[
a
,
c
]
,
P
)
;
∀
P
⊂
P
∗
}
,
B
=
{
S
_
(
f
/
[
c
,
b
]
,
P
)
;
∀
P
⊂
P
∗
}
{\displaystyle A=\{{\underline {S}}(f/[a,c],P);\forall P\subset P^{*}\},B=\{{\underline {S}}(f/[c,b],P);\forall P\subset P^{*}\}}
A
+
B
=
{
S
_
(
f
/
[
a
,
c
]
,
P
)
+
S
_
(
f
/
[
c
,
b
]
,
P
)
;
∀
P
⊂
P
∗
}
=
{
S
_
(
f
/
[
a
,
b
]
,
Q
)
;
∀
Q
⊂
Q
∗
}
{\displaystyle A+B=\{{\underline {S}}(f/[a,c],P)+{\underline {S}}(f/[c,b],P);\forall P\subset P^{*}\}=\{{\underline {S}}(f/[a,b],Q);\forall Q\subset Q^{*}\}}
pelo lema 2 e pelo lema 3 temos
∫
_
a
b
f
(
x
)
d
x
=
s
u
p
(
A
+
B
)
=
s
u
p
(
A
)
+
s
u
p
(
B
)
=
∫
_
a
c
f
(
x
)
d
x
+
∫
_
c
b
f
(
x
)
d
x
{\displaystyle {\underline {\int }}_{a}^{b}f(x)dx=sup(A+B)=sup(A)+sup(B)={\underline {\int }}_{a}^{c}f(x)dx+{\underline {\int }}_{c}^{b}f(x)dx}
(b)Sejam
A
=
{
S
¯
(
f
/
[
a
,
c
]
,
P
)
;
∀
P
⊂
P
∗
}
,
B
=
{
S
¯
(
f
/
[
c
,
b
]
,
P
)
;
∀
P
⊂
P
∗
}
{\displaystyle A=\{{\overline {S}}(f/[a,c],P);\forall P\subset P^{*}\},B=\{{\overline {S}}(f/[c,b],P);\forall P\subset P^{*}\}}
A
+
B
=
{
S
¯
(
f
/
[
a
,
c
]
,
P
)
+
S
¯
(
f
/
[
c
,
b
]
,
P
)
;
∀
P
⊂
P
∗
}
=
{
S
¯
(
f
/
[
a
,
b
]
,
Q
)
;
∀
Q
⊂
Q
∗
}
{\displaystyle A+B=\{{\overline {S}}(f/[a,c],P)+{\overline {S}}(f/[c,b],P);\forall P\subset P^{*}\}=\{{\overline {S}}(f/[a,b],Q);\forall Q\subset Q^{*}\}}
pelo lema 2 e pelo lema 3 temos
∫
¯
a
b
f
(
x
)
d
x
=
i
n
f
(
A
+
B
)
=
i
n
f
(
A
)
+
i
n
f
(
B
)
=
∫
¯
a
c
f
(
x
)
d
x
+
∫
¯
c
b
f
(
x
)
d
x
{\displaystyle {\overline {\int }}_{a}^{b}f(x)dx=inf(A+B)=inf(A)+inf(B)={\overline {\int }}_{a}^{c}f(x)dx+{\overline {\int }}_{c}^{b}f(x)dx}
Seja
A
′
⊂
R
{\displaystyle A'\subset \mathbb {R} }
A
=
{
x
∈
A
′
;
M
≤
x
≤
N
}
;
A
∩
[
M
,
N
]
≠
∅
{\displaystyle A=\{x\in A';M\leq x\leq N\};A\cap [M,N]\neq \emptyset }
c
∈
R
{\displaystyle c\in \mathbb {R} }
(a)Se c> 0, então
c
⋅
A
=
{
c
.
x
∈
A
′
;
c
⋅
M
≤
c
⋅
x
≤
c
⋅
N
}
{\displaystyle c\cdot A=\{c.x\in A';c\cdot M\leq c\cdot x\leq c\cdot N\}}
Assim:
s
u
p
(
c
⋅
A
)
=
c
⋅
s
u
p
(
A
)
e
i
n
f
(
c
⋅
A
)
=
c
⋅
i
n
f
(
A
)
{\displaystyle sup(c\cdot A)=c\cdot sup(A)\;e\;inf(c\cdot A)=c\cdot inf(A)}
(b)Se c< 0, então
c
⋅
A
=
{
c
.
x
∈
A
′
;
c
⋅
M
≥
c
⋅
x
≥
c
⋅
N
}
{\displaystyle c\cdot A=\{c.x\in A';c\cdot M\geq c\cdot x\geq c\cdot N\}}
Assim:
s
u
p
(
c
⋅
A
)
=
c
⋅
i
n
f
(
A
)
e
i
n
f
(
c
⋅
A
)
=
c
⋅
s
u
p
(
A
)
{\displaystyle sup(c\cdot A)=c\cdot inf(A)\;e\;inf(c\cdot A)=c\cdot sup(A)}
Demonstração
Editar
(a)
s
u
p
(
c
⋅
A
)
=
c
⋅
N
=
c
⋅
s
u
p
(
A
)
{\displaystyle sup(c\cdot A)=c\cdot N=c\cdot sup(A)}
i
n
f
(
c
⋅
A
)
=
c
⋅
M
=
c
⋅
i
n
f
(
A
)
{\displaystyle inf(c\cdot A)=c\cdot M=c\cdot inf(A)}
(b)
s
u
p
(
c
⋅
A
)
=
c
⋅
M
=
c
⋅
i
n
f
(
A
)
{\displaystyle sup(c\cdot A)=c\cdot M=c\cdot inf(A)}
i
n
f
(
c
⋅
A
)
=
c
⋅
N
=
c
⋅
s
u
p
(
A
)
{\displaystyle inf(c\cdot A)=c\cdot N=c\cdot sup(A)}
Sejam
f
,
g
:
[
a
,
b
]
↦
{\displaystyle f,g:[a,b]\mapsto }
(a)
∫
_
a
b
f
(
x
)
d
x
+
∫
_
a
b
g
(
x
)
d
x
≤
∫
_
a
b
[
f
(
x
)
+
g
(
x
)
]
d
x
≤
∫
¯
a
b
[
f
(
x
)
+
g
(
x
)
]
d
x
≤
∫
¯
a
b
f
(
x
)
d
x
+
∫
¯
a
b
g
(
x
)
d
x
{\displaystyle {\underline {\int }}_{a}^{b}f(x)dx+{\underline {\int }}_{a}^{b}g(x)dx\leq {\underline {\int }}_{a}^{b}[f(x)+g(x)]dx\leq {\overline {\int }}_{a}^{b}[f(x)+g(x)]dx\leq {\overline {\int }}_{a}^{b}f(x)dx+{\overline {\int }}_{a}^{b}g(x)dx}
(b)
c>0
∫
_
a
b
c
⋅
f
(
x
)
d
x
=
c
∫
_
a
b
f
(
x
)
d
x
{\displaystyle {\underline {\int }}_{a}^{b}c\cdot f(x)dx=c{\underline {\int }}_{a}^{b}f(x)dx}
∫
¯
a
b
c
⋅
f
(
x
)
d
x
=
c
∫
¯
a
b
f
(
x
)
d
x
{\displaystyle {\overline {\int }}_{a}^{b}c\cdot f(x)dx=c{\overline {\int }}_{a}^{b}f(x)dx}
c<0
∫
_
a
b
c
⋅
f
(
x
)
d
x
=
c
∫
¯
a
b
f
(
x
)
d
x
{\displaystyle {\underline {\int }}_{a}^{b}c\cdot f(x)dx=c{\overline {\int }}_{a}^{b}f(x)dx}
∫
¯
a
b
c
⋅
f
(
x
)
d
x
=
c
∫
_
a
b
f
(
x
)
d
x
{\displaystyle {\overline {\int }}_{a}^{b}c\cdot f(x)dx=c{\underline {\int }}_{a}^{b}f(x)dx}
(c)
S
e
f
(
x
)
<
g
(
x
)
∀
x
∈
[
a
,
b
]
{\displaystyle Sef(x)<g(x)\forall ~x\in [a,b]}
\underline {\int}_{a}^{b} f(x)dx \le \underline {\int}_{a}^{b} g(x)dx
\overline {\int}_{a}^{b} f(x)dx \le \overline {\int}_{a}^{b} g(x)dx Demonstração
Editar
Das somas de Darboux destacamos as seguintes propriedades elementares .
Para quaisquer partições
P
,
Q
∈
P
(
[
a
,
b
]
)
{\displaystyle P,Q\in {\mathcal {P}}([a,b])}
s
f
(
P
)
≤
S
f
(
Q
)
{\displaystyle s_{f}(P)\leq S_{f}(Q)}
Se
P
,
Q
∈
P
(
[
a
,
b
]
)
{\displaystyle P,Q\in {\mathcal {P}}([a,b])}
são duas partições tais que
P
⊂
Q
{\displaystyle P\subset Q}
(caso em que
Q
{\displaystyle Q}
P
{\displaystyle P}
P
{\displaystyle P}
então
s
f
(
P
)
≤
s
f
(
Q
)
{\displaystyle s_{f}(P)\leq s_{f}(Q)}
e
S
f
(
Q
)
≤
S
f
(
P
)
{\displaystyle S_{f}(Q)\leq S_{f}(P)}
As duas propriedades simples acima permitem-nos obter com facilidade a seguinte primeira condição de integrabilidade (ver [4]
Uma função
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\rightarrow \mathbb {R} }
Para cada
ϵ
>
0
{\displaystyle \epsilon >0}
P
∈
P
(
[
a
,
b
]
)
{\displaystyle P\in {\mathcal {P}}([a,b])}
S
f
(
P
)
−
s
f
(
P
)
<
ϵ
{\displaystyle S_{f}(P)-s_{f}(P)<\epsilon }
.Para provar este teorema comecemos por observar que pelas propriedades algébricas dos ínfimos e dos supremos se tem
∫
¯
a
b
f
(
x
)
d
x
−
∫
_
a
b
f
(
x
)
d
=
inf
{
S
f
(
Q
1
)
:
Q
1
∈
P
(
[
a
,
b
]
)
}
−
sup
{
s
f
(
Q
2
)
:
Q
2
∈
P
(
[
a
,
b
]
)
}
=
inf
{
S
f
(
Q
1
)
:
Q
1
∈
P
(
[
a
,
b
]
)
}
+
inf
{
−
s
f
(
Q
2
)
:
Q
2
∈
P
(
[
a
,
b
]
)
}
=
inf
{
S
f
(
Q
1
)
−
s
f
(
Q
2
)
:
Q
1
,
Q
2
∈
P
(
[
a
,
b
]
)
}
.
{\displaystyle {\begin{aligned}{\overline {\int }}_{a}^{\ b}f(x)\,\mathrm {d} x-{\underline {\int }}_{a}^{\ b}f(x)\,\mathrm {d} &=\inf \ \{S_{f}(Q_{1}):Q_{1}\in {\mathcal {P}}([a,b])\}-\sup\{s_{f}(Q_{2}):Q_{2}\in {\mathcal {P([a,b])}}\}\\&=\inf \ \{S_{f}(Q_{1}):Q_{1}\in {\mathcal {P}}([a,b])\}+\inf\{-s_{f}(Q_{2}):Q_{2}\in {\mathcal {P([a,b])}}\}\\&\\&=\inf \ \{S_{f}(Q_{1})-s_{f}(Q_{2}):Q_{1},Q_{2}\in {\mathcal {P}}([a,b])\}.\\\end{aligned}}}
Deste modo, se
f
{\displaystyle f}
ϵ
>
0
{\displaystyle \epsilon >0}
Q
1
,
Q
2
∈
P
(
[
a
,
b
]
)
{\displaystyle Q_{1},Q_{2}\in {\mathcal {P}}([a,b])}
0
≤
S
f
(
Q
1
)
−
s
f
(
Q
2
)
<
ϵ
{\displaystyle 0\leq S_{f}(Q_{1})-s_{f}(Q_{2})<\epsilon }
P
=
Q
1
∪
Q
2
{\displaystyle P=Q_{1}\cup Q_{2}}
0
≤
S
f
(
P
)
−
s
f
(
P
)
<
ϵ
.
{\displaystyle 0\leq S_{f}(P)-s_{f}(P)<\epsilon .}
Reciprocamente, se para cada
ϵ
>
0
{\displaystyle \epsilon >0}
P
∈
P
(
[
a
,
b
]
)
{\displaystyle P\in {\mathcal {P}}([a,b])}
0
≤
S
f
(
P
)
−
s
f
(
P
)
<
ϵ
.
{\displaystyle 0\leq S_{f}(P)-s_{f}(P)<\epsilon .}
0
≤
S
f
(
Q
1
)
−
s
f
(
Q
2
)
<
ϵ
{\displaystyle 0\leq S_{f}(Q_{1})-s_{f}(Q_{2})<\epsilon }
Q
1
,
Q
2
∈
P
(
[
a
,
b
]
)
{\displaystyle Q_{1},Q_{2}\in {\mathcal {P}}([a,b])}
P
.
{\displaystyle P.}
Exemplo 1 (funções monótonas)
Editar
Seja
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\rightarrow \mathbb {R} }
[
a
,
b
]
.
{\displaystyle [a,b].}
f
{\displaystyle f}
[
a
,
b
]
.
{\displaystyle [a,b].}
Supondo,por exemplo, que
f
{\displaystyle f}
[
a
,
b
]
{\displaystyle [a,b]}
−
f
{\displaystyle -f}
f
{\displaystyle f}
f
(
a
)
≤
f
/
x
)
≤
f
(
b
)
{\displaystyle f(a)\leq f/x)\leq f(b)}
x
∈
[
a
,
b
]
{\displaystyle x\in [a,b]}
P
=
{
x
0
,
x
1
,
.
.
.
,
x
n
}
{\displaystyle P=\{x_{0},x_{1},...,x_{n}\}}
[
a
,
b
]
{\displaystyle [a,b]}
S
f
(
P
)
−
s
f
(
P
)
=
∑
1
=
1
n
(
f
(
x
i
)
−
f
(
x
i
−
1
]
)
)
(
x
i
−
x
i
−
1
)
{\textstyle S_{f}(P)-s_{f}(P)=\sum _{1=1}^{n}(f(x_{i})-f(x_{i-1]}))(x_{i}-x_{i-1})}
(
x
i
−
x
i
−
1
)
≤
|
P
|
{\displaystyle (x_{i}-x_{i-1})\leq |P|}
S
f
(
P
)
−
s
f
(
P
)
≤
(
f
(
b
)
−
f
(
a
)
)
|
P
|
{\textstyle S_{f}(P)-s_{f}(P)\leq (f(b)-f(a))|P|}
ϵ
>
0
{\displaystyle \epsilon >0}
P
{\displaystyle P}
|
P
|
<
ϵ
/
(
f
(
b
)
−
f
(
a
)
)
{\displaystyle |P|<\epsilon /(f(b)-f(a))}
S
f
(
P
)
−
s
f
(
P
)
<
ϵ
{\displaystyle S_{f}(P)-s_{f}(P)<\epsilon }
f
{\displaystyle f}
[
a
,
b
]
.
{\displaystyle [a,b].}
Igualmente como aplicação da condição de Riemann podemos obter a integrabilidade das funções contínuas. Para o efeito, vamos usar uma propriedade importante das funções contínuas em intervalos compactos (isto é, fechados e limitados): a de serem uniformemente contínua s. Significa isto, que para qualquer
α
>
0
{\displaystyle \alpha >0}
β
>
0
{\displaystyle \beta >0}
|
f
(
x
)
−
f
(
y
)
|
<
α
{\textstyle |f(x)-f(y)|<\alpha }
|
x
−
y
|
<
β
{\displaystyle |x-y|<\beta }
Exemplo 2 (funções contínuas)
Editar
Seja
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\rightarrow \mathbb {R} }
[
a
,
b
]
.
{\displaystyle [a,b].}
f
{\displaystyle f}
[
a
,
b
]
.
{\displaystyle [a,b].}
Comecemos por notar que, pelo teorema de Weierstrass ,
f
{\displaystyle f}
[
a
,
b
]
{\displaystyle [a,b]}
i
=
1
,
.
.
.
,
n
,
{\displaystyle i=1,...,n,}
m
i
=
f
(
c
i
)
{\displaystyle m_{i}=f(c_{i})}
M
i
=
f
(
d
i
)
{\displaystyle M_{i}=f(d_{i})}
c
i
,
d
i
∈
[
x
i
−
1
,
x
i
]
{\displaystyle c_{i},d_{i}\in [x_{i-1},x_{i}]}
P
=
{
x
0
,
x
1
,
.
.
.
,
x
n
}
{\displaystyle P=\{x_{0},x_{1},...,x_{n}\}}
[
a
,
b
]
{\displaystyle [a,b]}
S
f
(
P
)
−
s
f
(
P
)
=
∑
i
=
1
n
(
f
(
d
i
)
−
f
(
c
i
)
)
(
x
i
−
x
i
−
1
)
{\textstyle S_{f}(P)-s_{f}(P)=\sum _{i=1}^{n}(f(d_{i})-f(c_{i}))(x_{i}-x_{i-1})}
Então na condição de Riemann tomemos
ϵ
>
0
{\displaystyle \epsilon >0}
f
{\displaystyle f}
[
a
,
b
]
{\displaystyle [a,b]}
α
=
ϵ
/
(
b
−
a
)
{\displaystyle \alpha =\epsilon /(b-a)}
β
>
0
{\textstyle \beta >0}
P
{\displaystyle P}
[
a
,
b
]
{\displaystyle [a,b]}
|
P
|
<
β
{\displaystyle |P|<\beta }
i
=
1
,
.
.
.
,
n
,
{\displaystyle i=1,...,n,}
0
≤
(
f
(
d
i
)
−
f
(
c
i
)
)
<
ϵ
/
(
b
−
a
)
{\textstyle 0\leq (f(d_{i})-f(c_{i}))<\epsilon /(b-a)}
S
f
(
P
)
−
s
f
(
P
)
<
ϵ
.
{\textstyle S_{f}(P)-s_{f}(P)<\epsilon .}
f
{\displaystyle f}
[
a
,
b
]
.
{\displaystyle [a,b].}
A condição do teorema assume um aspeto meramente técnico. Ela não nos dá qualquer indício das qualidades que a função deva verificar para ser integrável à Riemann.Um quadro qualitativo desta propriedade, aparece pela mão de Henri Lebesgue, na sua tese doutoral ("Intégrale, Longueur, Aire" (Integral, Comprimento, Área) apresentada na Faculdade de Ciências de Paris em 1902, com base no conceito de conjunto de medida de nula.
![{\displaystyle [t_{0},t_{1}]{\mbox{ e }}[t_{1},t_{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/819bb6cfc1e812c411c6e416932af29fdb7ab71a)
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