Proof of $fin L^1(mathbb R)$ then $m{xmid |f|=infty }=0$. Is my proof correct.Proving that $int 0 dmu=0$An...
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Proof of $fin L^1(mathbb R)$ then $m{xmid |f|=infty }=0$. Is my proof correct.
Proving that $int 0 dmu=0$An application of the General Lebesgue Dominated convergence theoremIf $f in L_p cap L_q$, $1 leq p < r < q < infty$ then $f in L_r$.Suppose $mu$ is a finite measure and $sup_n int |f_n|^{1+epsilon} dmu<infty$ for some $epsilon$. Prove that ${f_n}$ is uniformly integrableFatou Lemma extension for every measurable sequence (negative & non-negative)Real Analysis, Folland The Dominated Convergence Theorem$f_n rightarrow f$ a.e and $int_mathbb{R} f_n dm rightarrow int_mathbb{R} f dm$ implies $int_E f_n dm rightarrow int_E f dm$Show that $g$ is integrable and that $int_E f,dmu = int_E g ,dmu, ,,Einmathcal{A}.$Fatou's Lemma proof from Royden 4thProblems proving that if $f_nrightarrow f$ pointwise and $int_R f=lim_{n}int_R f_n$ then $int_E f=lim_{n}int_E f_n$ for meas $E subseteq R$.why $f$ and $g$ need to be bounded
$begingroup$
Let $fin L^1(mathbb R)$. Then $f$ is finite a.e. I did the following proof and my teacher gave me a mark of $0$. What I did is : Let $E={xmid |f(x)|=infty }$. We have that$$ int_E|f|leq int_{mathbb R}|f|$$
If $m(E)>0$ then $$int_E|f|=infty cdot m(E)=infty,$$
and thus $int_{mathbb R}|f|=infty $ which is a contradiction.
He says that $acdot infty =infty $ for $a>0$ is more a convention than something formal. After he justify that if my proof is valid, then someone could make the proof $$infty cdot m(E)=int_E|f|leq int_{mathbb R}|f|<infty ,$$
and thus $m(E)=0$. Which is not completely wrong, but with rigor $infty cdot 0$ is undeterminated. So he didn't accepted my proof.
I'm a bit confuse because for me it's a complete valid proof. What do you think ?
measure-theory lebesgue-integral
asked Mar 20 at 12:49
PierrePierre
19011
$endgroup$
add a comment |
$begingroup$
Let $fin L^1(mathbb R)$. Then $f$ is finite a.e. I did the following proof and my teacher gave me a mark of $0$. What I did is : Let $E={xmid |f(x)|=infty }$. We have that$$ int_E|f|leq int_{mathbb R}|f|$$
If $m(E)>0$ then $$int_E|f|=infty cdot m(E)=infty,$$
and thus $int_{mathbb R}|f|=infty $ which is a contradiction.
He says that $acdot infty =infty $ for $a>0$ is more a convention than something formal. After he justify that if my proof is valid, then someone could make the proof $$infty cdot m(E)=int_E|f|leq int_{mathbb R}|f|<infty ,$$
and thus $m(E)=0$. Which is not completely wrong, but with rigor $infty cdot 0$ is undeterminated. So he didn't accepted my proof.
I'm a bit confuse because for me it's a complete valid proof. What do you think ?
measure-theory lebesgue-integral
asked Mar 20 at 12:49
PierrePierre
19011
$endgroup$
$begingroup$
Relevant: math.stackexchange.com/q/1514331/66096
$endgroup$
– Gabriel Romon
Mar 20 at 13:02
1
$begingroup$
Also for me it is a complete valid proof. Nothing is wrong with defining a multiplication on $[0,infty]$ where $rcdotinfty:=infty$ if $r>0$ and $0cdotinfty=0$. And this multiplication can be applied by measure theory. If that is not okay according to your teacher then at least he is obliged to show where this leads to wrong conclusions. He did not manage to do that in his "justification". If such multiplication is defined then $inftycdot0$ is no longer indeterminate.
$endgroup$
– drhab
Mar 20 at 13:07
$begingroup$
FWIW: Rudin's 1.22, "Arithmetic in $[0,infty]$," of Real and Complex Analysis, has "Let us define " and then writes down the arithmetic (including $inftycdot 0= 0$ which you are willing to consider as problematic). On the other hand, he did write "let us define," which is your teacher's 'convention' point, I suppose. I think what you had to prove was obvious - and your proof boils down to "it's obvious". And that is the pedagogic/legal problem... My opinion is that your teacher erred in the formulation of the question.
$endgroup$
– peter a g
Mar 20 at 13:13
$begingroup$
BTW, Rudin points out that with $inftycdot 0 = 0$, the usual laws of distributivity associativity etc hold in $[0,infty]$.
$endgroup$
– peter a g
Mar 20 at 13:16
add a comment |
$begingroup$
Let $fin L^1(mathbb R)$. Then $f$ is finite a.e. I did the following proof and my teacher gave me a mark of $0$. What I did is : Let $E={xmid |f(x)|=infty }$. We have that$$ int_E|f|leq int_{mathbb R}|f|$$
If $m(E)>0$ then $$int_E|f|=infty cdot m(E)=infty,$$
and thus $int_{mathbb R}|f|=infty $ which is a contradiction.
He says that $acdot infty =infty $ for $a>0$ is more a convention than something formal. After he justify that if my proof is valid, then someone could make the proof $$infty cdot m(E)=int_E|f|leq int_{mathbb R}|f|<infty ,$$
and thus $m(E)=0$. Which is not completely wrong, but with rigor $infty cdot 0$ is undeterminated. So he didn't accepted my proof.
I'm a bit confuse because for me it's a complete valid proof. What do you think ?
measure-theory lebesgue-integral
asked Mar 20 at 12:49
PierrePierre
19011
$endgroup$
Let $fin L^1(mathbb R)$. Then $f$ is finite a.e. I did the following proof and my teacher gave me a mark of $0$. What I did is : Let $E={xmid |f(x)|=infty }$. We have that$$ int_E|f|leq int_{mathbb R}|f|$$
If $m(E)>0$ then $$int_E|f|=infty cdot m(E)=infty,$$
and thus $int_{mathbb R}|f|=infty $ which is a contradiction.
He says that $acdot infty =infty $ for $a>0$ is more a convention than something formal. After he justify that if my proof is valid, then someone could make the proof $$infty cdot m(E)=int_E|f|leq int_{mathbb R}|f|<infty ,$$
and thus $m(E)=0$. Which is not completely wrong, but with rigor $infty cdot 0$ is undeterminated. So he didn't accepted my proof.
I'm a bit confuse because for me it's a complete valid proof. What do you think ?
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Mar 20 at 12:49
PierrePierre
19011
asked Mar 20 at 12:49
PierrePierre
19011
asked Mar 20 at 12:49
PierrePierre
19011
asked Mar 20 at 12:49
PierrePierre
19011
asked Mar 20 at 12:49
PierrePierre
19011
19011
$begingroup$
Relevant: math.stackexchange.com/q/1514331/66096
$endgroup$
– Gabriel Romon
Mar 20 at 13:02
1
$begingroup$
Also for me it is a complete valid proof. Nothing is wrong with defining a multiplication on $[0,infty]$ where $rcdotinfty:=infty$ if $r>0$ and $0cdotinfty=0$. And this multiplication can be applied by measure theory. If that is not okay according to your teacher then at least he is obliged to show where this leads to wrong conclusions. He did not manage to do that in his "justification". If such multiplication is defined then $inftycdot0$ is no longer indeterminate.
$endgroup$
– drhab
Mar 20 at 13:07
$begingroup$
FWIW: Rudin's 1.22, "Arithmetic in $[0,infty]$," of Real and Complex Analysis, has "Let us define " and then writes down the arithmetic (including $inftycdot 0= 0$ which you are willing to consider as problematic). On the other hand, he did write "let us define," which is your teacher's 'convention' point, I suppose. I think what you had to prove was obvious - and your proof boils down to "it's obvious". And that is the pedagogic/legal problem... My opinion is that your teacher erred in the formulation of the question.
$endgroup$
– peter a g
Mar 20 at 13:13
$begingroup$
BTW, Rudin points out that with $inftycdot 0 = 0$, the usual laws of distributivity associativity etc hold in $[0,infty]$.
$endgroup$
– peter a g
Mar 20 at 13:16
add a comment |
$begingroup$
Relevant: math.stackexchange.com/q/1514331/66096
$endgroup$
– Gabriel Romon
Mar 20 at 13:02
1
$begingroup$
Also for me it is a complete valid proof. Nothing is wrong with defining a multiplication on $[0,infty]$ where $rcdotinfty:=infty$ if $r>0$ and $0cdotinfty=0$. And this multiplication can be applied by measure theory. If that is not okay according to your teacher then at least he is obliged to show where this leads to wrong conclusions. He did not manage to do that in his "justification". If such multiplication is defined then $inftycdot0$ is no longer indeterminate.
$endgroup$
– drhab
Mar 20 at 13:07
$begingroup$
FWIW: Rudin's 1.22, "Arithmetic in $[0,infty]$," of Real and Complex Analysis, has "Let us define " and then writes down the arithmetic (including $inftycdot 0= 0$ which you are willing to consider as problematic). On the other hand, he did write "let us define," which is your teacher's 'convention' point, I suppose. I think what you had to prove was obvious - and your proof boils down to "it's obvious". And that is the pedagogic/legal problem... My opinion is that your teacher erred in the formulation of the question.
$endgroup$
– peter a g
Mar 20 at 13:13
$begingroup$
BTW, Rudin points out that with $inftycdot 0 = 0$, the usual laws of distributivity associativity etc hold in $[0,infty]$.
$endgroup$
– peter a g
Mar 20 at 13:16
$begingroup$
Relevant: math.stackexchange.com/q/1514331/66096
$endgroup$
– Gabriel Romon
Mar 20 at 13:02
$begingroup$
Relevant: math.stackexchange.com/q/1514331/66096
$endgroup$
– Gabriel Romon
Mar 20 at 13:02
1
1
$begingroup$
Also for me it is a complete valid proof. Nothing is wrong with defining a multiplication on $[0,infty]$ where $rcdotinfty:=infty$ if $r>0$ and $0cdotinfty=0$. And this multiplication can be applied by measure theory. If that is not okay according to your teacher then at least he is obliged to show where this leads to wrong conclusions. He did not manage to do that in his "justification". If such multiplication is defined then $inftycdot0$ is no longer indeterminate.
$endgroup$
– drhab
Mar 20 at 13:07
$begingroup$
Also for me it is a complete valid proof. Nothing is wrong with defining a multiplication on $[0,infty]$ where $rcdotinfty:=infty$ if $r>0$ and $0cdotinfty=0$. And this multiplication can be applied by measure theory. If that is not okay according to your teacher then at least he is obliged to show where this leads to wrong conclusions. He did not manage to do that in his "justification". If such multiplication is defined then $inftycdot0$ is no longer indeterminate.
$endgroup$
– drhab
Mar 20 at 13:07
$begingroup$
FWIW: Rudin's 1.22, "Arithmetic in $[0,infty]$," of Real and Complex Analysis, has "Let us define " and then writes down the arithmetic (including $inftycdot 0= 0$ which you are willing to consider as problematic). On the other hand, he did write "let us define," which is your teacher's 'convention' point, I suppose. I think what you had to prove was obvious - and your proof boils down to "it's obvious". And that is the pedagogic/legal problem... My opinion is that your teacher erred in the formulation of the question.
$endgroup$
– peter a g
Mar 20 at 13:13
$begingroup$
FWIW: Rudin's 1.22, "Arithmetic in $[0,infty]$," of Real and Complex Analysis, has "Let us define " and then writes down the arithmetic (including $inftycdot 0= 0$ which you are willing to consider as problematic). On the other hand, he did write "let us define," which is your teacher's 'convention' point, I suppose. I think what you had to prove was obvious - and your proof boils down to "it's obvious". And that is the pedagogic/legal problem... My opinion is that your teacher erred in the formulation of the question.
$endgroup$
– peter a g
Mar 20 at 13:13
$begingroup$
BTW, Rudin points out that with $inftycdot 0 = 0$, the usual laws of distributivity associativity etc hold in $[0,infty]$.
$endgroup$
– peter a g
Mar 20 at 13:16
$begingroup$
BTW, Rudin points out that with $inftycdot 0 = 0$, the usual laws of distributivity associativity etc hold in $[0,infty]$.
$endgroup$
– peter a g
Mar 20 at 13:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your argument can be made rigorous, but when you observe conventions for multiplication with $infty$ it should be understood in terms of limits.
Assume for simplicity that $f ge 0$ and that $E = {f = infty}$. For all $n ge 1$ you have $n chi_E le f$ so by monotonicity of the integral
$$n cdot m(E) = int_{mathbf R} n chi_E le int_{mathbf R}f$$
In the limit you obtain (formally) $infty cdot m(E) le displaystyle int_{mathbf R} f < infty$, but the actual contradiction follows from the observation that if $m(E) > 0$ you may select $$n > frac 1{m(E)} int_{mathbf R} f.$$
answered Mar 20 at 15:01
Umberto P.Umberto P.
40.3k13370
$endgroup$
add a comment |
$begingroup$
Firstly, since $infty$ is not a number, $inftycdot |E|$ does not make sense if $|E|=0$. For example,
$$ ncdotfrac1n=1$$
which leads to
$$ inftycdot0=1.$$
Secondly, you assume $|E|>0$ to get
$$ |E|lefrac1{infty}int_R|f|dx=0 $$
and hence $|E|=0$ which is against your assumption $|E|>0$.
To void this, you can define
$$ E_n={xin R: |f(x)|ge n} $$
and then
$$ lim E_n=E. $$
Since
$$ int_R|f(x)|dx ge int_{E_n}|f(x)|dxge n|E_n|$$
one has
$$ |E_n|le frac1n int_R|f(x)|dx.$$
So
$$ lim |E_n|=0 $$
or
$$ |E|=0.$$
answered Mar 20 at 13:51
xpaulxpaul
23.5k24655
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
Your argument can be made rigorous, but when you observe conventions for multiplication with $infty$ it should be understood in terms of limits.
Assume for simplicity that $f ge 0$ and that $E = {f = infty}$. For all $n ge 1$ you have $n chi_E le f$ so by monotonicity of the integral
$$n cdot m(E) = int_{mathbf R} n chi_E le int_{mathbf R}f$$
In the limit you obtain (formally) $infty cdot m(E) le displaystyle int_{mathbf R} f < infty$, but the actual contradiction follows from the observation that if $m(E) > 0$ you may select $$n > frac 1{m(E)} int_{mathbf R} f.$$
answered Mar 20 at 15:01
Umberto P.Umberto P.
40.3k13370
$endgroup$
add a comment |
$begingroup$
Your argument can be made rigorous, but when you observe conventions for multiplication with $infty$ it should be understood in terms of limits.
Assume for simplicity that $f ge 0$ and that $E = {f = infty}$. For all $n ge 1$ you have $n chi_E le f$ so by monotonicity of the integral
$$n cdot m(E) = int_{mathbf R} n chi_E le int_{mathbf R}f$$
In the limit you obtain (formally) $infty cdot m(E) le displaystyle int_{mathbf R} f < infty$, but the actual contradiction follows from the observation that if $m(E) > 0$ you may select $$n > frac 1{m(E)} int_{mathbf R} f.$$
answered Mar 20 at 15:01
Umberto P.Umberto P.
40.3k13370
$endgroup$
add a comment |
$begingroup$
Your argument can be made rigorous, but when you observe conventions for multiplication with $infty$ it should be understood in terms of limits.
Assume for simplicity that $f ge 0$ and that $E = {f = infty}$. For all $n ge 1$ you have $n chi_E le f$ so by monotonicity of the integral
$$n cdot m(E) = int_{mathbf R} n chi_E le int_{mathbf R}f$$
In the limit you obtain (formally) $infty cdot m(E) le displaystyle int_{mathbf R} f < infty$, but the actual contradiction follows from the observation that if $m(E) > 0$ you may select $$n > frac 1{m(E)} int_{mathbf R} f.$$
answered Mar 20 at 15:01
Umberto P.Umberto P.
40.3k13370
$endgroup$
Your argument can be made rigorous, but when you observe conventions for multiplication with $infty$ it should be understood in terms of limits.
Assume for simplicity that $f ge 0$ and that $E = {f = infty}$. For all $n ge 1$ you have $n chi_E le f$ so by monotonicity of the integral
$$n cdot m(E) = int_{mathbf R} n chi_E le int_{mathbf R}f$$
In the limit you obtain (formally) $infty cdot m(E) le displaystyle int_{mathbf R} f < infty$, but the actual contradiction follows from the observation that if $m(E) > 0$ you may select $$n > frac 1{m(E)} int_{mathbf R} f.$$
answered Mar 20 at 15:01
Umberto P.Umberto P.
40.3k13370
answered Mar 20 at 15:01
Umberto P.Umberto P.
40.3k13370
answered Mar 20 at 15:01
Umberto P.Umberto P.
40.3k13370
answered Mar 20 at 15:01
Umberto P.Umberto P.
40.3k13370
40.3k13370
add a comment |
add a comment |
$begingroup$
Firstly, since $infty$ is not a number, $inftycdot |E|$ does not make sense if $|E|=0$. For example,
$$ ncdotfrac1n=1$$
which leads to
$$ inftycdot0=1.$$
Secondly, you assume $|E|>0$ to get
$$ |E|lefrac1{infty}int_R|f|dx=0 $$
and hence $|E|=0$ which is against your assumption $|E|>0$.
To void this, you can define
$$ E_n={xin R: |f(x)|ge n} $$
and then
$$ lim E_n=E. $$
Since
$$ int_R|f(x)|dx ge int_{E_n}|f(x)|dxge n|E_n|$$
one has
$$ |E_n|le frac1n int_R|f(x)|dx.$$
So
$$ lim |E_n|=0 $$
or
$$ |E|=0.$$
answered Mar 20 at 13:51
xpaulxpaul
23.5k24655
$endgroup$
add a comment |
$begingroup$
Firstly, since $infty$ is not a number, $inftycdot |E|$ does not make sense if $|E|=0$. For example,
$$ ncdotfrac1n=1$$
which leads to
$$ inftycdot0=1.$$
Secondly, you assume $|E|>0$ to get
$$ |E|lefrac1{infty}int_R|f|dx=0 $$
and hence $|E|=0$ which is against your assumption $|E|>0$.
To void this, you can define
$$ E_n={xin R: |f(x)|ge n} $$
and then
$$ lim E_n=E. $$
Since
$$ int_R|f(x)|dx ge int_{E_n}|f(x)|dxge n|E_n|$$
one has
$$ |E_n|le frac1n int_R|f(x)|dx.$$
So
$$ lim |E_n|=0 $$
or
$$ |E|=0.$$
answered Mar 20 at 13:51
xpaulxpaul
23.5k24655
$endgroup$
add a comment |
$begingroup$
Firstly, since $infty$ is not a number, $inftycdot |E|$ does not make sense if $|E|=0$. For example,
$$ ncdotfrac1n=1$$
which leads to
$$ inftycdot0=1.$$
Secondly, you assume $|E|>0$ to get
$$ |E|lefrac1{infty}int_R|f|dx=0 $$
and hence $|E|=0$ which is against your assumption $|E|>0$.
To void this, you can define
$$ E_n={xin R: |f(x)|ge n} $$
and then
$$ lim E_n=E. $$
Since
$$ int_R|f(x)|dx ge int_{E_n}|f(x)|dxge n|E_n|$$
one has
$$ |E_n|le frac1n int_R|f(x)|dx.$$
So
$$ lim |E_n|=0 $$
or
$$ |E|=0.$$
answered Mar 20 at 13:51
xpaulxpaul
23.5k24655
$endgroup$
Firstly, since $infty$ is not a number, $inftycdot |E|$ does not make sense if $|E|=0$. For example,
$$ ncdotfrac1n=1$$
which leads to
$$ inftycdot0=1.$$
Secondly, you assume $|E|>0$ to get
$$ |E|lefrac1{infty}int_R|f|dx=0 $$
and hence $|E|=0$ which is against your assumption $|E|>0$.
To void this, you can define
$$ E_n={xin R: |f(x)|ge n} $$
and then
$$ lim E_n=E. $$
Since
$$ int_R|f(x)|dx ge int_{E_n}|f(x)|dxge n|E_n|$$
one has
$$ |E_n|le frac1n int_R|f(x)|dx.$$
So
$$ lim |E_n|=0 $$
or
$$ |E|=0.$$
answered Mar 20 at 13:51
xpaulxpaul
23.5k24655
edited Mar 20 at 15:07
answered Mar 20 at 13:51
xpaulxpaul
23.5k24655
answered Mar 20 at 13:51
xpaulxpaul
23.5k24655
answered Mar 20 at 13:51
xpaulxpaul
23.5k24655
23.5k24655
add a comment |
add a comment |
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Relevant: math.stackexchange.com/q/1514331/66096
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– Gabriel Romon
Mar 20 at 13:02
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Also for me it is a complete valid proof. Nothing is wrong with defining a multiplication on $[0,infty]$ where $rcdotinfty:=infty$ if $r>0$ and $0cdotinfty=0$. And this multiplication can be applied by measure theory. If that is not okay according to your teacher then at least he is obliged to show where this leads to wrong conclusions. He did not manage to do that in his "justification". If such multiplication is defined then $inftycdot0$ is no longer indeterminate.
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– drhab
Mar 20 at 13:07
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FWIW: Rudin's 1.22, "Arithmetic in $[0,infty]$," of Real and Complex Analysis, has "Let us define " and then writes down the arithmetic (including $inftycdot 0= 0$ which you are willing to consider as problematic). On the other hand, he did write "let us define," which is your teacher's 'convention' point, I suppose. I think what you had to prove was obvious - and your proof boils down to "it's obvious". And that is the pedagogic/legal problem... My opinion is that your teacher erred in the formulation of the question.
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– peter a g
Mar 20 at 13:13
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BTW, Rudin points out that with $inftycdot 0 = 0$, the usual laws of distributivity associativity etc hold in $[0,infty]$.
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– peter a g
Mar 20 at 13:16