Let $C in mathbbR^n$ be a Jordan mensurable set of measure zero. Show that $int_A chi_C = 0$, for every closed rectangle $A supset C$.

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Let $C in mathbbR^n$ be a Jordan mensurable set of measure zero. Show that $int_A chi_C = 0$, for every closed rectangle $A supset C$.

A subset $A$ of $mathbbR^n$ has measure 0 if for every $epsilon > 0$ there is a cover $U_1, U_2,... $ of A by closed rectangles such that $sum_i=1^inftyv(U_i) < epsilon.$

A bounded set C is called Jordan-measurable if for every closed rectangle A such that $C subset A$ we have that $chi_C: A to mathbbR$ (given by $chi_C(x) = 1$ if $x in C$, and $chi_C(x) = 0$ if $x in A-C$) is integrable. In this case $v(C) = int_A chi_C$

---

The only thing I know is that by hypothesis we have that $int_A chi_C = v(C)$. I am really bad in analysis.

Any help would be great.

integration analysis

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edited Jul 15 at 2:48

asked Jul 15 at 2:23

P.G

15910

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up vote
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Let $C in mathbbR^n$ be a Jordan mensurable set of measure zero. Show that $int_A chi_C = 0$, for every closed rectangle $A supset C$.

A subset $A$ of $mathbbR^n$ has measure 0 if for every $epsilon > 0$ there is a cover $U_1, U_2,... $ of A by closed rectangles such that $sum_i=1^inftyv(U_i) < epsilon.$

A bounded set C is called Jordan-measurable if for every closed rectangle A such that $C subset A$ we have that $chi_C: A to mathbbR$ (given by $chi_C(x) = 1$ if $x in C$, and $chi_C(x) = 0$ if $x in A-C$) is integrable. In this case $v(C) = int_A chi_C$

---

The only thing I know is that by hypothesis we have that $int_A chi_C = v(C)$. I am really bad in analysis.

Any help would be great.

integration analysis

share|cite|improve this question

edited Jul 15 at 2:48

asked Jul 15 at 2:23

P.G

15910

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $C in mathbbR^n$ be a Jordan mensurable set of measure zero. Show that $int_A chi_C = 0$, for every closed rectangle $A supset C$.

A subset $A$ of $mathbbR^n$ has measure 0 if for every $epsilon > 0$ there is a cover $U_1, U_2,... $ of A by closed rectangles such that $sum_i=1^inftyv(U_i) < epsilon.$

A bounded set C is called Jordan-measurable if for every closed rectangle A such that $C subset A$ we have that $chi_C: A to mathbbR$ (given by $chi_C(x) = 1$ if $x in C$, and $chi_C(x) = 0$ if $x in A-C$) is integrable. In this case $v(C) = int_A chi_C$

---

The only thing I know is that by hypothesis we have that $int_A chi_C = v(C)$. I am really bad in analysis.

Any help would be great.

integration analysis

share|cite|improve this question

edited Jul 15 at 2:48

asked Jul 15 at 2:23

P.G

15910

Let $C in mathbbR^n$ be a Jordan mensurable set of measure zero. Show that $int_A chi_C = 0$, for every closed rectangle $A supset C$.

A subset $A$ of $mathbbR^n$ has measure 0 if for every $epsilon > 0$ there is a cover $U_1, U_2,... $ of A by closed rectangles such that $sum_i=1^inftyv(U_i) < epsilon.$

A bounded set C is called Jordan-measurable if for every closed rectangle A such that $C subset A$ we have that $chi_C: A to mathbbR$ (given by $chi_C(x) = 1$ if $x in C$, and $chi_C(x) = 0$ if $x in A-C$) is integrable. In this case $v(C) = int_A chi_C$

---

The only thing I know is that by hypothesis we have that $int_A chi_C = v(C)$. I am really bad in analysis.

Any help would be great.

integration analysis

share|cite|improve this question

edited Jul 15 at 2:48

asked Jul 15 at 2:23

P.G

15910

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 2:48

edited Jul 15 at 2:48

edited Jul 15 at 2:48

asked Jul 15 at 2:23

P.G

15910

asked Jul 15 at 2:23

P.G

15910

asked Jul 15 at 2:23

P.G

15910

15910

add a comment | 

add a comment | 

1 Answer
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We can write A as $Ccup (A - C)$ and we can express the integral as $int_A chi_C = int_C chi_C + int_A-Cchi_C$.

By the definition of $chi_C$, we have that $chi_C=0$ on $A-C$, so the second integral of the sum is zero. Then $int_A chi_C = int_C chi_C = v(C)$. Which is zero because C has measure zero (or alternatively we can show by the definition that $v(C)$ is less than $epsilon$ for every $epsilon > 0$

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answered Jul 15 at 4:21

ricardorr

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1 Answer
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votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

We can write A as $Ccup (A - C)$ and we can express the integral as $int_A chi_C = int_C chi_C + int_A-Cchi_C$.

By the definition of $chi_C$, we have that $chi_C=0$ on $A-C$, so the second integral of the sum is zero. Then $int_A chi_C = int_C chi_C = v(C)$. Which is zero because C has measure zero (or alternatively we can show by the definition that $v(C)$ is less than $epsilon$ for every $epsilon > 0$

share|cite|improve this answer

answered Jul 15 at 4:21

ricardorr

364

add a comment | 

up vote
2
down vote



accepted

We can write A as $Ccup (A - C)$ and we can express the integral as $int_A chi_C = int_C chi_C + int_A-Cchi_C$.

By the definition of $chi_C$, we have that $chi_C=0$ on $A-C$, so the second integral of the sum is zero. Then $int_A chi_C = int_C chi_C = v(C)$. Which is zero because C has measure zero (or alternatively we can show by the definition that $v(C)$ is less than $epsilon$ for every $epsilon > 0$

share|cite|improve this answer

answered Jul 15 at 4:21

ricardorr

364

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

We can write A as $Ccup (A - C)$ and we can express the integral as $int_A chi_C = int_C chi_C + int_A-Cchi_C$.

By the definition of $chi_C$, we have that $chi_C=0$ on $A-C$, so the second integral of the sum is zero. Then $int_A chi_C = int_C chi_C = v(C)$. Which is zero because C has measure zero (or alternatively we can show by the definition that $v(C)$ is less than $epsilon$ for every $epsilon > 0$

share|cite|improve this answer

answered Jul 15 at 4:21

ricardorr

364

We can write A as $Ccup (A - C)$ and we can express the integral as $int_A chi_C = int_C chi_C + int_A-Cchi_C$.

By the definition of $chi_C$, we have that $chi_C=0$ on $A-C$, so the second integral of the sum is zero. Then $int_A chi_C = int_C chi_C = v(C)$. Which is zero because C has measure zero (or alternatively we can show by the definition that $v(C)$ is less than $epsilon$ for every $epsilon > 0$

share|cite|improve this answer

answered Jul 15 at 4:21

ricardorr

364

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 4:21

ricardorr

364

answered Jul 15 at 4:21

ricardorr

364

answered Jul 15 at 4:21

ricardorr

364

364

add a comment | 

add a comment | 

 

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