Proving Tchebychev's Inequalityexistence of Lebesgue integralMarkov inequality limitSuppose $mu$ is a finite...
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Proving Tchebychev's Inequality
existence of Lebesgue integralMarkov inequality limitSuppose $mu$ is a finite measure and $sup_n int |f_n|^{1+epsilon} dmu<infty$ for some $epsilon$. Prove that ${f_n}$ is uniformly integrableProve that $lim_{tto infty} tmu({x:f(x)geq t})=0$Strengthened Chebyshev InequalityLebesgue integral of f over R is bigger or equal to alpha times measure of $E_alpha?$Improper integral $int_0^infty cos(x^2)$ exists but $cos(x^2)$ is not Lebesgue integrableProving Fatou's lemma from the DCTProving the Borel-Cantelli LemmaProving $g_ain L^1(mathbb{R}^d) iff a<-d$
$begingroup$
Suppose $fgeq 0$, and $f$ is lebesgue integrable. If $alpha>0$ and $E_alpha={x:f(x)>alpha}$, prove that
$$ m(E_alpha)leq 1/alpha int f $$
My proof attempt
Proof. Let the assumptions be as above. Then
$$ alpha 1_{E_alpha}<f(x)1_{E_alpha} $$
Then by monotonicity of the integral
$$ alphaint 1_{E_alpha}<int f(x)1_{E_alpha}leq int f(x) $$
$implies m(E_alpha)<1/alphaint f$.
My main concern is that I end up with $<$ instead of $leq.$ Any feedback on the proof or style is much appreciated.
integration measure-theory proof-verification lebesgue-integral
asked Mar 12 at 14:57
Joe Man AnalysisJoe Man Analysis
56419
$endgroup$
add a comment |
$begingroup$
Suppose $fgeq 0$, and $f$ is lebesgue integrable. If $alpha>0$ and $E_alpha={x:f(x)>alpha}$, prove that
$$ m(E_alpha)leq 1/alpha int f $$
My proof attempt
Proof. Let the assumptions be as above. Then
$$ alpha 1_{E_alpha}<f(x)1_{E_alpha} $$
Then by monotonicity of the integral
$$ alphaint 1_{E_alpha}<int f(x)1_{E_alpha}leq int f(x) $$
$implies m(E_alpha)<1/alphaint f$.
My main concern is that I end up with $<$ instead of $leq.$ Any feedback on the proof or style is much appreciated.
integration measure-theory proof-verification lebesgue-integral
asked Mar 12 at 14:57
Joe Man AnalysisJoe Man Analysis
56419
$endgroup$
add a comment |
$begingroup$
Suppose $fgeq 0$, and $f$ is lebesgue integrable. If $alpha>0$ and $E_alpha={x:f(x)>alpha}$, prove that
$$ m(E_alpha)leq 1/alpha int f $$
My proof attempt
Proof. Let the assumptions be as above. Then
$$ alpha 1_{E_alpha}<f(x)1_{E_alpha} $$
Then by monotonicity of the integral
$$ alphaint 1_{E_alpha}<int f(x)1_{E_alpha}leq int f(x) $$
$implies m(E_alpha)<1/alphaint f$.
My main concern is that I end up with $<$ instead of $leq.$ Any feedback on the proof or style is much appreciated.
integration measure-theory proof-verification lebesgue-integral
asked Mar 12 at 14:57
Joe Man AnalysisJoe Man Analysis
56419
$endgroup$
Suppose $fgeq 0$, and $f$ is lebesgue integrable. If $alpha>0$ and $E_alpha={x:f(x)>alpha}$, prove that
$$ m(E_alpha)leq 1/alpha int f $$
My proof attempt
Proof. Let the assumptions be as above. Then
$$ alpha 1_{E_alpha}<f(x)1_{E_alpha} $$
Then by monotonicity of the integral
$$ alphaint 1_{E_alpha}<int f(x)1_{E_alpha}leq int f(x) $$
$implies m(E_alpha)<1/alphaint f$.
My main concern is that I end up with $<$ instead of $leq.$ Any feedback on the proof or style is much appreciated.
integration measure-theory proof-verification lebesgue-integral
integration measure-theory proof-verification lebesgue-integral
asked Mar 12 at 14:57
Joe Man AnalysisJoe Man Analysis
56419
asked Mar 12 at 14:57
Joe Man AnalysisJoe Man Analysis
56419
asked Mar 12 at 14:57
Joe Man AnalysisJoe Man Analysis
56419
asked Mar 12 at 14:57
Joe Man AnalysisJoe Man Analysis
56419
asked Mar 12 at 14:57
Joe Man AnalysisJoe Man Analysis
56419
56419
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Firstly you have not specified what $x$ is, and note that $alpha mathbf1_{E_alpha}<f(x)mathbf1_{E_alpha}$ is not true for all $x$. What we have is that $alphamathbf1_{E_alpha}(x)<f(x)mathbf1_{E_alpha}(x)$ for all $xin E_alpha$. For all $xnotin E_alpha$ we have $alphamathbf1_{E_alpha}(x)=0=f(x)mathbf1_{E_alpha}(x)$. Thus what we have in general is that $amathbf1_{E_alpha}leq fmathbf1_{E_alpha}$. From this point on your proof seems fine.
answered Mar 12 at 17:26
K.PowerK.Power
3,490926
$endgroup$
$begingroup$
Ah, thank you! That makes a lot of sense
$endgroup$
– Joe Man Analysis
Mar 12 at 17:31
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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$begingroup$
Firstly you have not specified what $x$ is, and note that $alpha mathbf1_{E_alpha}<f(x)mathbf1_{E_alpha}$ is not true for all $x$. What we have is that $alphamathbf1_{E_alpha}(x)<f(x)mathbf1_{E_alpha}(x)$ for all $xin E_alpha$. For all $xnotin E_alpha$ we have $alphamathbf1_{E_alpha}(x)=0=f(x)mathbf1_{E_alpha}(x)$. Thus what we have in general is that $amathbf1_{E_alpha}leq fmathbf1_{E_alpha}$. From this point on your proof seems fine.
answered Mar 12 at 17:26
K.PowerK.Power
3,490926
$endgroup$
$begingroup$
Ah, thank you! That makes a lot of sense
$endgroup$
– Joe Man Analysis
Mar 12 at 17:31
add a comment |
$begingroup$
Firstly you have not specified what $x$ is, and note that $alpha mathbf1_{E_alpha}<f(x)mathbf1_{E_alpha}$ is not true for all $x$. What we have is that $alphamathbf1_{E_alpha}(x)<f(x)mathbf1_{E_alpha}(x)$ for all $xin E_alpha$. For all $xnotin E_alpha$ we have $alphamathbf1_{E_alpha}(x)=0=f(x)mathbf1_{E_alpha}(x)$. Thus what we have in general is that $amathbf1_{E_alpha}leq fmathbf1_{E_alpha}$. From this point on your proof seems fine.
answered Mar 12 at 17:26
K.PowerK.Power
3,490926
$endgroup$
$begingroup$
Ah, thank you! That makes a lot of sense
$endgroup$
– Joe Man Analysis
Mar 12 at 17:31
add a comment |
$begingroup$
Firstly you have not specified what $x$ is, and note that $alpha mathbf1_{E_alpha}<f(x)mathbf1_{E_alpha}$ is not true for all $x$. What we have is that $alphamathbf1_{E_alpha}(x)<f(x)mathbf1_{E_alpha}(x)$ for all $xin E_alpha$. For all $xnotin E_alpha$ we have $alphamathbf1_{E_alpha}(x)=0=f(x)mathbf1_{E_alpha}(x)$. Thus what we have in general is that $amathbf1_{E_alpha}leq fmathbf1_{E_alpha}$. From this point on your proof seems fine.
answered Mar 12 at 17:26
K.PowerK.Power
3,490926
$endgroup$
Firstly you have not specified what $x$ is, and note that $alpha mathbf1_{E_alpha}<f(x)mathbf1_{E_alpha}$ is not true for all $x$. What we have is that $alphamathbf1_{E_alpha}(x)<f(x)mathbf1_{E_alpha}(x)$ for all $xin E_alpha$. For all $xnotin E_alpha$ we have $alphamathbf1_{E_alpha}(x)=0=f(x)mathbf1_{E_alpha}(x)$. Thus what we have in general is that $amathbf1_{E_alpha}leq fmathbf1_{E_alpha}$. From this point on your proof seems fine.
answered Mar 12 at 17:26
K.PowerK.Power
3,490926
answered Mar 12 at 17:26
K.PowerK.Power
3,490926
answered Mar 12 at 17:26
K.PowerK.Power
3,490926
answered Mar 12 at 17:26
K.PowerK.Power
3,490926
3,490926
$begingroup$
Ah, thank you! That makes a lot of sense
$endgroup$
– Joe Man Analysis
Mar 12 at 17:31
add a comment |
$begingroup$
Ah, thank you! That makes a lot of sense
$endgroup$
– Joe Man Analysis
Mar 12 at 17:31
$begingroup$
Ah, thank you! That makes a lot of sense
$endgroup$
– Joe Man Analysis
Mar 12 at 17:31
$begingroup$
Ah, thank you! That makes a lot of sense
$endgroup$
– Joe Man Analysis
Mar 12 at 17:31
add a comment |
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