Rudin real anlysis: prove that $f$ is Riemann integrable
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Let $f : [0,1] →Bbb R$ satisfy the property that for every $ε>0$ the set ${x∈[0,1]: |f(x)|ge ε}$ is finite. Prove that for every continuous increasing function $α :[0,1]→Bbb R$, that $f$ is Riemann integral and integral $int_0^1f,dα=0$.
I am confused about what is the meaning of the the set ${x∈[0,1]: |f(x)|ge ε}$ is finite.
And I want to apply theorem 6.10(rudin):suppose f is bounded on[a,b], f has only finitely many points of discontinuity on[a,b], and α is continuous at every point at which f is discontinuous. Then f is Riemann integral.
real-analysis integration riemann-integration
asked Dec 9 '18 at 4:14
BearryBearry
112
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add a comment |
$begingroup$
Let $f : [0,1] →Bbb R$ satisfy the property that for every $ε>0$ the set ${x∈[0,1]: |f(x)|ge ε}$ is finite. Prove that for every continuous increasing function $α :[0,1]→Bbb R$, that $f$ is Riemann integral and integral $int_0^1f,dα=0$.
I am confused about what is the meaning of the the set ${x∈[0,1]: |f(x)|ge ε}$ is finite.
And I want to apply theorem 6.10(rudin):suppose f is bounded on[a,b], f has only finitely many points of discontinuity on[a,b], and α is continuous at every point at which f is discontinuous. Then f is Riemann integral.
real-analysis integration riemann-integration
asked Dec 9 '18 at 4:14
BearryBearry
112
$endgroup$
2
$begingroup$
in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
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– Masacroso
Dec 9 '18 at 4:20
$begingroup$
The set is finite, i.e. the set contains only finitely many elements.
$endgroup$
– xbh
Dec 9 '18 at 5:37
$begingroup$
Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
$endgroup$
– copper.hat
Dec 9 '18 at 5:42
add a comment |
$begingroup$
Let $f : [0,1] →Bbb R$ satisfy the property that for every $ε>0$ the set ${x∈[0,1]: |f(x)|ge ε}$ is finite. Prove that for every continuous increasing function $α :[0,1]→Bbb R$, that $f$ is Riemann integral and integral $int_0^1f,dα=0$.
I am confused about what is the meaning of the the set ${x∈[0,1]: |f(x)|ge ε}$ is finite.
And I want to apply theorem 6.10(rudin):suppose f is bounded on[a,b], f has only finitely many points of discontinuity on[a,b], and α is continuous at every point at which f is discontinuous. Then f is Riemann integral.
real-analysis integration riemann-integration
asked Dec 9 '18 at 4:14
BearryBearry
112
$endgroup$
Let $f : [0,1] →Bbb R$ satisfy the property that for every $ε>0$ the set ${x∈[0,1]: |f(x)|ge ε}$ is finite. Prove that for every continuous increasing function $α :[0,1]→Bbb R$, that $f$ is Riemann integral and integral $int_0^1f,dα=0$.
I am confused about what is the meaning of the the set ${x∈[0,1]: |f(x)|ge ε}$ is finite.
And I want to apply theorem 6.10(rudin):suppose f is bounded on[a,b], f has only finitely many points of discontinuity on[a,b], and α is continuous at every point at which f is discontinuous. Then f is Riemann integral.
real-analysis integration riemann-integration
real-analysis integration riemann-integration
asked Dec 9 '18 at 4:14
BearryBearry
112
asked Dec 9 '18 at 4:14
BearryBearry
112
edited Dec 9 '18 at 4:28
Bearry
asked Dec 9 '18 at 4:14
BearryBearry
112
asked Dec 9 '18 at 4:14
BearryBearry
112
asked Dec 9 '18 at 4:14
BearryBearry
112
112
2
$begingroup$
in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
$endgroup$
– Masacroso
Dec 9 '18 at 4:20
$begingroup$
The set is finite, i.e. the set contains only finitely many elements.
$endgroup$
– xbh
Dec 9 '18 at 5:37
$begingroup$
Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
$endgroup$
– copper.hat
Dec 9 '18 at 5:42
add a comment |
2
$begingroup$
in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
$endgroup$
– Masacroso
Dec 9 '18 at 4:20
$begingroup$
The set is finite, i.e. the set contains only finitely many elements.
$endgroup$
– xbh
Dec 9 '18 at 5:37
$begingroup$
Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
$endgroup$
– copper.hat
Dec 9 '18 at 5:42
2
2
$begingroup$
in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
$endgroup$
– Masacroso
Dec 9 '18 at 4:20
$begingroup$
in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
$endgroup$
– Masacroso
Dec 9 '18 at 4:20
$begingroup$
The set is finite, i.e. the set contains only finitely many elements.
$endgroup$
– xbh
Dec 9 '18 at 5:37
$begingroup$
The set is finite, i.e. the set contains only finitely many elements.
$endgroup$
– xbh
Dec 9 '18 at 5:37
$begingroup$
Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
$endgroup$
– copper.hat
Dec 9 '18 at 5:42
$begingroup$
Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
$endgroup$
– copper.hat
Dec 9 '18 at 5:42
add a comment |
1 Answer
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$begingroup$
First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] to mathbf R$ defined by
$$ g(x) = begin{cases} 0 & x notin mathbf Q \ frac{1}{q} & x = frac pq text{ where } gcd(p,q) = 1 end{cases} $$
Notice that there are only finitely many rational numbers with denominator $le N$ for any given integer $N$.
On the other hand, let
$$ A = {x : f(x) ne 0} = bigcup_{n ge 1} left{x in [0,1] : |f(x)| ge frac1n right} $$
You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)
For $alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.
Here's the sketch when $alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $alpha$.
There are finitely many points, say $x_1,dots,x_N$ in $[0,1]$ where $|f(x)| > delta$. Surround these points $x_1,dots,x_N$ by small intervals, $I_1,dots,I_N$. Let $J = [0,1] setminus (I_1 cup cdots cup I_N)$. Then a Riemann sum over this partition is bounded by
$$ left(sup_J |f(x)|right) operatorname{length}(J) + left(sup_{I_1} |f(x)|right) operatorname{length}(I_1) + dots + left(sup_{I_N} |f(x)|right) operatorname{length}(I_N) $$
If you play with the numbers right you can get this $< varepsilon$.
answered Dec 9 '18 at 6:41
Trevor GunnTrevor Gunn
14.4k32046
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add a comment |
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$begingroup$
First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] to mathbf R$ defined by
$$ g(x) = begin{cases} 0 & x notin mathbf Q \ frac{1}{q} & x = frac pq text{ where } gcd(p,q) = 1 end{cases} $$
Notice that there are only finitely many rational numbers with denominator $le N$ for any given integer $N$.
On the other hand, let
$$ A = {x : f(x) ne 0} = bigcup_{n ge 1} left{x in [0,1] : |f(x)| ge frac1n right} $$
You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)
For $alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.
Here's the sketch when $alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $alpha$.
There are finitely many points, say $x_1,dots,x_N$ in $[0,1]$ where $|f(x)| > delta$. Surround these points $x_1,dots,x_N$ by small intervals, $I_1,dots,I_N$. Let $J = [0,1] setminus (I_1 cup cdots cup I_N)$. Then a Riemann sum over this partition is bounded by
$$ left(sup_J |f(x)|right) operatorname{length}(J) + left(sup_{I_1} |f(x)|right) operatorname{length}(I_1) + dots + left(sup_{I_N} |f(x)|right) operatorname{length}(I_N) $$
If you play with the numbers right you can get this $< varepsilon$.
answered Dec 9 '18 at 6:41
Trevor GunnTrevor Gunn
14.4k32046
$endgroup$
add a comment |
$begingroup$
First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] to mathbf R$ defined by
$$ g(x) = begin{cases} 0 & x notin mathbf Q \ frac{1}{q} & x = frac pq text{ where } gcd(p,q) = 1 end{cases} $$
Notice that there are only finitely many rational numbers with denominator $le N$ for any given integer $N$.
On the other hand, let
$$ A = {x : f(x) ne 0} = bigcup_{n ge 1} left{x in [0,1] : |f(x)| ge frac1n right} $$
You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)
For $alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.
Here's the sketch when $alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $alpha$.
There are finitely many points, say $x_1,dots,x_N$ in $[0,1]$ where $|f(x)| > delta$. Surround these points $x_1,dots,x_N$ by small intervals, $I_1,dots,I_N$. Let $J = [0,1] setminus (I_1 cup cdots cup I_N)$. Then a Riemann sum over this partition is bounded by
$$ left(sup_J |f(x)|right) operatorname{length}(J) + left(sup_{I_1} |f(x)|right) operatorname{length}(I_1) + dots + left(sup_{I_N} |f(x)|right) operatorname{length}(I_N) $$
If you play with the numbers right you can get this $< varepsilon$.
answered Dec 9 '18 at 6:41
Trevor GunnTrevor Gunn
14.4k32046
$endgroup$
add a comment |
$begingroup$
First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] to mathbf R$ defined by
$$ g(x) = begin{cases} 0 & x notin mathbf Q \ frac{1}{q} & x = frac pq text{ where } gcd(p,q) = 1 end{cases} $$
Notice that there are only finitely many rational numbers with denominator $le N$ for any given integer $N$.
On the other hand, let
$$ A = {x : f(x) ne 0} = bigcup_{n ge 1} left{x in [0,1] : |f(x)| ge frac1n right} $$
You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)
For $alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.
Here's the sketch when $alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $alpha$.
There are finitely many points, say $x_1,dots,x_N$ in $[0,1]$ where $|f(x)| > delta$. Surround these points $x_1,dots,x_N$ by small intervals, $I_1,dots,I_N$. Let $J = [0,1] setminus (I_1 cup cdots cup I_N)$. Then a Riemann sum over this partition is bounded by
$$ left(sup_J |f(x)|right) operatorname{length}(J) + left(sup_{I_1} |f(x)|right) operatorname{length}(I_1) + dots + left(sup_{I_N} |f(x)|right) operatorname{length}(I_N) $$
If you play with the numbers right you can get this $< varepsilon$.
answered Dec 9 '18 at 6:41
Trevor GunnTrevor Gunn
14.4k32046
$endgroup$
First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] to mathbf R$ defined by
$$ g(x) = begin{cases} 0 & x notin mathbf Q \ frac{1}{q} & x = frac pq text{ where } gcd(p,q) = 1 end{cases} $$
Notice that there are only finitely many rational numbers with denominator $le N$ for any given integer $N$.
On the other hand, let
$$ A = {x : f(x) ne 0} = bigcup_{n ge 1} left{x in [0,1] : |f(x)| ge frac1n right} $$
You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)
For $alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.
Here's the sketch when $alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $alpha$.
There are finitely many points, say $x_1,dots,x_N$ in $[0,1]$ where $|f(x)| > delta$. Surround these points $x_1,dots,x_N$ by small intervals, $I_1,dots,I_N$. Let $J = [0,1] setminus (I_1 cup cdots cup I_N)$. Then a Riemann sum over this partition is bounded by
$$ left(sup_J |f(x)|right) operatorname{length}(J) + left(sup_{I_1} |f(x)|right) operatorname{length}(I_1) + dots + left(sup_{I_N} |f(x)|right) operatorname{length}(I_N) $$
If you play with the numbers right you can get this $< varepsilon$.
answered Dec 9 '18 at 6:41
Trevor GunnTrevor Gunn
14.4k32046
answered Dec 9 '18 at 6:41
Trevor GunnTrevor Gunn
14.4k32046
answered Dec 9 '18 at 6:41
Trevor GunnTrevor Gunn
14.4k32046
answered Dec 9 '18 at 6:41
Trevor GunnTrevor Gunn
14.4k32046
14.4k32046
add a comment |
add a comment |
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$begingroup$
in this site the questions that doesn't show some effort to solve it are not well-received. So I recommend to you that edit your question showing what had you tried to solve the posted exercise.
$endgroup$
– Masacroso
Dec 9 '18 at 4:20
$begingroup$
The set is finite, i.e. the set contains only finitely many elements.
$endgroup$
– xbh
Dec 9 '18 at 5:37
$begingroup$
Loosely, it means that if you pick an $epsilon>0$ then the number of points at which $f(x)$ lies outside the band $[-epsilon,epsilon]$ is finite, and it is straightforward to create a partition that essentially 'ignores' these finite points. In particular, the lower and upper sums for the integral will be bounded below by $-epsilon(alpha(1)-alpha(0))$ and above by $epsilon (alpha(1)-alpha(0))$. Hence the integral will be zero. Continuity of $alpha$ is used to 'ignore' the finite points.
$endgroup$
– copper.hat
Dec 9 '18 at 5:42