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BASIC REAL AND COMPLEX ANALYSIS
Integrals
Lebesgue measure
ovolssnul
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ovollb2lem
Metamath Proof Explorer
Ascii
Unicode
Theorem
ovolssnul
Description:
A subset of a nullset is null.
(Contributed by
Mario Carneiro
, 19-Mar-2014)
Ref
Expression
Assertion
ovolssnul
⊢
A
⊆
B
∧
B
⊆
ℝ
∧
vol
*
⁡
B
=
0
→
vol
*
⁡
A
=
0
Proof
Step
Hyp
Ref
Expression
1
ovolss
⊢
A
⊆
B
∧
B
⊆
ℝ
→
vol
*
⁡
A
≤
vol
*
⁡
B
2
1
3adant3
⊢
A
⊆
B
∧
B
⊆
ℝ
∧
vol
*
⁡
B
=
0
→
vol
*
⁡
A
≤
vol
*
⁡
B
3
simp3
⊢
A
⊆
B
∧
B
⊆
ℝ
∧
vol
*
⁡
B
=
0
→
vol
*
⁡
B
=
0
4
2
3
breqtrd
⊢
A
⊆
B
∧
B
⊆
ℝ
∧
vol
*
⁡
B
=
0
→
vol
*
⁡
A
≤
0
5
sstr
⊢
A
⊆
B
∧
B
⊆
ℝ
→
A
⊆
ℝ
6
5
3adant3
⊢
A
⊆
B
∧
B
⊆
ℝ
∧
vol
*
⁡
B
=
0
→
A
⊆
ℝ
7
ovolge0
⊢
A
⊆
ℝ
→
0
≤
vol
*
⁡
A
8
6
7
syl
⊢
A
⊆
B
∧
B
⊆
ℝ
∧
vol
*
⁡
B
=
0
→
0
≤
vol
*
⁡
A
9
ovolcl
⊢
A
⊆
ℝ
→
vol
*
⁡
A
∈
ℝ
*
10
6
9
syl
⊢
A
⊆
B
∧
B
⊆
ℝ
∧
vol
*
⁡
B
=
0
→
vol
*
⁡
A
∈
ℝ
*
11
0xr
⊢
0
∈
ℝ
*
12
xrletri3
⊢
vol
*
⁡
A
∈
ℝ
*
∧
0
∈
ℝ
*
→
vol
*
⁡
A
=
0
↔
vol
*
⁡
A
≤
0
∧
0
≤
vol
*
⁡
A
13
10
11
12
sylancl
⊢
A
⊆
B
∧
B
⊆
ℝ
∧
vol
*
⁡
B
=
0
→
vol
*
⁡
A
=
0
↔
vol
*
⁡
A
≤
0
∧
0
≤
vol
*
⁡
A
14
4
8
13
mpbir2and
⊢
A
⊆
B
∧
B
⊆
ℝ
∧
vol
*
⁡
B
=
0
→
vol
*
⁡
A
=
0