Clarification on IntegrabilityProve this inequality $lvert a - brvert < frac12lvert b rvert implies lvert a rvert > frac12lvert b rvert$A Riemann integral with a jump discontinuity at its lower limitProve if $f(0) = 0$ then $lim_x to 0^+xint_x^1 fracf(t)t^2dt = 0$ for regulated function $f$Improper integral and its convergence. Is this procedure correct? Any quicker and simpler ways?Prob. 7 (b), Chap. 6, in Baby Rudin: Example of a function such that $lim_c to 0+ int_c^1 f(x) mathrmdx$ exists but . . .The old and modern definitions of total variation are actually equivalent?Prove that $a_n=fracnn+1$ is convergentPrinciples of math analysis by Rudin, Chapter 6 Problem 7Find the flaw in the given proof: about the limit of a sequence
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Clarification on Integrability
Prove this inequality $lvert a - brvert < frac12lvert b rvert implies lvert a rvert > frac12lvert b rvert$A Riemann integral with a jump discontinuity at its lower limitProve if $f(0) = 0$ then $lim_x to 0^+xint_x^1 fracf(t)t^2dt = 0$ for regulated function $f$Improper integral and its convergence. Is this procedure correct? Any quicker and simpler ways?Prob. 7 (b), Chap. 6, in Baby Rudin: Example of a function such that $lim_c to 0+ int_c^1 f(x) mathrmdx$ exists but . . .The old and modern definitions of total variation are actually equivalent?Prove that $a_n=fracnn+1$ is convergentPrinciples of math analysis by Rudin, Chapter 6 Problem 7Find the flaw in the given proof: about the limit of a sequence
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
If $f$ in integrable on some interval $[a,b]$ then we know that $lvert f rvert $ is also integrable on that same interval.
There is a problem in Rudin's Principles of Mathematical analysis such that we construct an $f$ where
$displaystyle int_0^1 f dx = lim_c downarrow 0 int_c^1 fdx$
exists and yet for $lvert f rvert$ this limit fails to exist.
How does this not contradict the implication above?
One such constuction is to set $f(x) = (-1)^k+1(k+1), forall x in (frac1k+1,frac1k]$.
real-analysis integration improper-integrals
asked 8 hours ago
all.overall.over
897 bronze badges
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add a comment |
$begingroup$
If $f$ in integrable on some interval $[a,b]$ then we know that $lvert f rvert $ is also integrable on that same interval.
There is a problem in Rudin's Principles of Mathematical analysis such that we construct an $f$ where
$displaystyle int_0^1 f dx = lim_c downarrow 0 int_c^1 fdx$
exists and yet for $lvert f rvert$ this limit fails to exist.
How does this not contradict the implication above?
One such constuction is to set $f(x) = (-1)^k+1(k+1), forall x in (frac1k+1,frac1k]$.
real-analysis integration improper-integrals
asked 8 hours ago
all.overall.over
897 bronze badges
$endgroup$
add a comment |
$begingroup$
If $f$ in integrable on some interval $[a,b]$ then we know that $lvert f rvert $ is also integrable on that same interval.
There is a problem in Rudin's Principles of Mathematical analysis such that we construct an $f$ where
$displaystyle int_0^1 f dx = lim_c downarrow 0 int_c^1 fdx$
exists and yet for $lvert f rvert$ this limit fails to exist.
How does this not contradict the implication above?
One such constuction is to set $f(x) = (-1)^k+1(k+1), forall x in (frac1k+1,frac1k]$.
real-analysis integration improper-integrals
asked 8 hours ago
all.overall.over
897 bronze badges
$endgroup$
If $f$ in integrable on some interval $[a,b]$ then we know that $lvert f rvert $ is also integrable on that same interval.
There is a problem in Rudin's Principles of Mathematical analysis such that we construct an $f$ where
$displaystyle int_0^1 f dx = lim_c downarrow 0 int_c^1 fdx$
exists and yet for $lvert f rvert$ this limit fails to exist.
How does this not contradict the implication above?
One such constuction is to set $f(x) = (-1)^k+1(k+1), forall x in (frac1k+1,frac1k]$.
real-analysis integration improper-integrals
real-analysis integration improper-integrals
asked 8 hours ago
all.overall.over
897 bronze badges
asked 8 hours ago
all.overall.over
897 bronze badges
asked 8 hours ago
all.overall.over
897 bronze badges
asked 8 hours ago
all.overall.over
897 bronze badges
asked 8 hours ago
all.overall.over
897 bronze badges
897 bronze badges
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2 Answers
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There is no contradiction because $f$ is not Riemann-integrable on $[0,1]$. The fact that the limit $lim_c to 0int_c ^1f(x)dx$ exists does not mean that $f$ is Riemann-integrable on $[0,1]$. Note that Rudin says in the exercise that we can define the symbol $int_0^1f(x)dx$ to mean said limit in the case where we have a function on $(0,1]$. It does not mean that $f$ is Riemann-integrable on $[0,1]$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.
PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $int_ 0^1f(x)dx$ agree when $f$ is Riemann-integrable.
answered 7 hours ago
Aloizio Macedo♦Aloizio Macedo
24.3k2 gold badges40 silver badges89 bronze badges
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The implication $f$ is integrable $Rightarrow$ $|f|$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.
answered 8 hours ago
Hans EnglerHans Engler
11k1 gold badge20 silver badges36 bronze badges
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4
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This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does.
$endgroup$
– Aloizio Macedo♦
7 hours ago
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
There is no contradiction because $f$ is not Riemann-integrable on $[0,1]$. The fact that the limit $lim_c to 0int_c ^1f(x)dx$ exists does not mean that $f$ is Riemann-integrable on $[0,1]$. Note that Rudin says in the exercise that we can define the symbol $int_0^1f(x)dx$ to mean said limit in the case where we have a function on $(0,1]$. It does not mean that $f$ is Riemann-integrable on $[0,1]$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.
PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $int_ 0^1f(x)dx$ agree when $f$ is Riemann-integrable.
answered 7 hours ago
Aloizio Macedo♦Aloizio Macedo
24.3k2 gold badges40 silver badges89 bronze badges
$endgroup$
add a comment |
$begingroup$
There is no contradiction because $f$ is not Riemann-integrable on $[0,1]$. The fact that the limit $lim_c to 0int_c ^1f(x)dx$ exists does not mean that $f$ is Riemann-integrable on $[0,1]$. Note that Rudin says in the exercise that we can define the symbol $int_0^1f(x)dx$ to mean said limit in the case where we have a function on $(0,1]$. It does not mean that $f$ is Riemann-integrable on $[0,1]$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.
PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $int_ 0^1f(x)dx$ agree when $f$ is Riemann-integrable.
answered 7 hours ago
Aloizio Macedo♦Aloizio Macedo
24.3k2 gold badges40 silver badges89 bronze badges
$endgroup$
add a comment |
$begingroup$
There is no contradiction because $f$ is not Riemann-integrable on $[0,1]$. The fact that the limit $lim_c to 0int_c ^1f(x)dx$ exists does not mean that $f$ is Riemann-integrable on $[0,1]$. Note that Rudin says in the exercise that we can define the symbol $int_0^1f(x)dx$ to mean said limit in the case where we have a function on $(0,1]$. It does not mean that $f$ is Riemann-integrable on $[0,1]$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.
PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $int_ 0^1f(x)dx$ agree when $f$ is Riemann-integrable.
answered 7 hours ago
Aloizio Macedo♦Aloizio Macedo
24.3k2 gold badges40 silver badges89 bronze badges
$endgroup$
There is no contradiction because $f$ is not Riemann-integrable on $[0,1]$. The fact that the limit $lim_c to 0int_c ^1f(x)dx$ exists does not mean that $f$ is Riemann-integrable on $[0,1]$. Note that Rudin says in the exercise that we can define the symbol $int_0^1f(x)dx$ to mean said limit in the case where we have a function on $(0,1]$. It does not mean that $f$ is Riemann-integrable on $[0,1]$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.
PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $int_ 0^1f(x)dx$ agree when $f$ is Riemann-integrable.
answered 7 hours ago
Aloizio Macedo♦Aloizio Macedo
24.3k2 gold badges40 silver badges89 bronze badges
edited 7 hours ago
answered 7 hours ago
Aloizio Macedo♦Aloizio Macedo
24.3k2 gold badges40 silver badges89 bronze badges
answered 7 hours ago
Aloizio Macedo♦Aloizio Macedo
24.3k2 gold badges40 silver badges89 bronze badges
answered 7 hours ago
Aloizio Macedo♦Aloizio Macedo
24.3k2 gold badges40 silver badges89 bronze badges
24.3k2 gold badges40 silver badges89 bronze badges
add a comment |
add a comment |
$begingroup$
The implication $f$ is integrable $Rightarrow$ $|f|$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.
answered 8 hours ago
Hans EnglerHans Engler
11k1 gold badge20 silver badges36 bronze badges
$endgroup$
4
$begingroup$
This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does.
$endgroup$
– Aloizio Macedo♦
7 hours ago
add a comment |
$begingroup$
The implication $f$ is integrable $Rightarrow$ $|f|$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.
answered 8 hours ago
Hans EnglerHans Engler
11k1 gold badge20 silver badges36 bronze badges
$endgroup$
4
$begingroup$
This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does.
$endgroup$
– Aloizio Macedo♦
7 hours ago
add a comment |
$begingroup$
The implication $f$ is integrable $Rightarrow$ $|f|$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.
answered 8 hours ago
Hans EnglerHans Engler
11k1 gold badge20 silver badges36 bronze badges
$endgroup$
The implication $f$ is integrable $Rightarrow$ $|f|$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.
answered 8 hours ago
Hans EnglerHans Engler
11k1 gold badge20 silver badges36 bronze badges
answered 8 hours ago
Hans EnglerHans Engler
11k1 gold badge20 silver badges36 bronze badges
answered 8 hours ago
Hans EnglerHans Engler
11k1 gold badge20 silver badges36 bronze badges
answered 8 hours ago
Hans EnglerHans Engler
11k1 gold badge20 silver badges36 bronze badges
11k1 gold badge20 silver badges36 bronze badges
4
$begingroup$
This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does.
$endgroup$
– Aloizio Macedo♦
7 hours ago
add a comment |
4
$begingroup$
This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does.
$endgroup$
– Aloizio Macedo♦
7 hours ago
4
4
$begingroup$
This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does.
$endgroup$
– Aloizio Macedo♦
7 hours ago
$begingroup$
This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does.
$endgroup$
– Aloizio Macedo♦
7 hours ago
add a comment |
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