Mixing
Let $f$ be Lebesgueable integrable, then $lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)| = 0$
$begingroup$
I must show that
Let $f$ be Lebesgueable integrable, then
$$lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy = 0$$
for almost every $x in mathbb{R}$.
Tentative proof:
$$|f(x) - f(y)| geq 0 Rightarrow lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy geq 0$$
We must show that
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy leq 0$$
I will show that for any $epsilon >0$
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy < epsilon$$
If $f(x) = 0$, By absolute continuity of integration, for any $epsilon >0$, $exists delta >0$ such that if
$$ m({|x-y| < h}) < delta Rightarrow int_{|x -y| < h}|f(y)|dy <frac{epsilon}{2} $$
Now $$ lim_{hto 0}sup m({|x-y| < h}) leq m({|x-y| < h}) < delta$$
then
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy < frac{epsilon}{2} < epsilon $$
Now for $f(x) neq 0$,
$$int_{|x -y| < h}|f(x) - f(y)|dy leq int_{|x -y| < h}|f(x)|dy + int_{|x -y| < h}|f(y)|dy $$
$$= |f(x)|m({|x -y| < h}) + int_{|x -y| < h}|f(y)|dy $$
and $|f(x)| < infty$ for allmost all $x in mathbb{R}$, since $f$ is Lebesgue integrable.
Let $delta^{*} < min {delta,frac{epsilon}{2|f(x)|}}$, now by absolute continuiuty of integration, for $epsilon > 0$,
$$m({|x -y| < h}) < delta^{*} < delta Rightarrow int_{|x -y| < h}|f(x) - f(y)|dy leq |f(x)|m({|x -y| < h}) + int_{|x -y| < h}|f(y)|dy < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$$
For almost every $x in mathbb{R}$. Hence,
$$ lim_{hto 0}sup m({|x-y| < h}) leq m({|x-y| < h}) < delta^{*} < delta Rightarrow lim_{hto 0}supint_{|x -y| < h}|f(x) - f(y)|dy < epsilon $$
Therefore,
$$lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy = 0$$
for almost every $x in mathbb{R}$.
lebesgue-integral lebesgue-measure limsup-and-liminf absolute-continuity
asked Jan 14 at 17:09
Richard ClareRichard Clare
1,076314
$endgroup$
add a comment |
$begingroup$
I must show that
Let $f$ be Lebesgueable integrable, then
$$lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy = 0$$
for almost every $x in mathbb{R}$.
Tentative proof:
$$|f(x) - f(y)| geq 0 Rightarrow lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy geq 0$$
We must show that
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy leq 0$$
I will show that for any $epsilon >0$
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy < epsilon$$
If $f(x) = 0$, By absolute continuity of integration, for any $epsilon >0$, $exists delta >0$ such that if
$$ m({|x-y| < h}) < delta Rightarrow int_{|x -y| < h}|f(y)|dy <frac{epsilon}{2} $$
Now $$ lim_{hto 0}sup m({|x-y| < h}) leq m({|x-y| < h}) < delta$$
then
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy < frac{epsilon}{2} < epsilon $$
Now for $f(x) neq 0$,
$$int_{|x -y| < h}|f(x) - f(y)|dy leq int_{|x -y| < h}|f(x)|dy + int_{|x -y| < h}|f(y)|dy $$
$$= |f(x)|m({|x -y| < h}) + int_{|x -y| < h}|f(y)|dy $$
and $|f(x)| < infty$ for allmost all $x in mathbb{R}$, since $f$ is Lebesgue integrable.
Let $delta^{*} < min {delta,frac{epsilon}{2|f(x)|}}$, now by absolute continuiuty of integration, for $epsilon > 0$,
$$m({|x -y| < h}) < delta^{*} < delta Rightarrow int_{|x -y| < h}|f(x) - f(y)|dy leq |f(x)|m({|x -y| < h}) + int_{|x -y| < h}|f(y)|dy < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$$
For almost every $x in mathbb{R}$. Hence,
$$ lim_{hto 0}sup m({|x-y| < h}) leq m({|x-y| < h}) < delta^{*} < delta Rightarrow lim_{hto 0}supint_{|x -y| < h}|f(x) - f(y)|dy < epsilon $$
Therefore,
$$lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy = 0$$
for almost every $x in mathbb{R}$.
lebesgue-integral lebesgue-measure limsup-and-liminf absolute-continuity
asked Jan 14 at 17:09
Richard ClareRichard Clare
1,076314
$endgroup$
1
$begingroup$
Sorry but is a bit complicate to read your proof (too long ^^). btw, it's not clear if you integrate wrt $x$, $y$ or both. Anyway, for me it's just monotone convergence with $g_n(x,y)=|f(x)-f(y)|boldsymbol 1_{{|x-y|<frac{1}{n}}}$. Should work, no ?
$endgroup$
– Surb
Jan 14 at 17:30
$begingroup$
its with respect to $y$. And I see what you mean, but that set is not increasing $|x-y| < 1/n$ is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:34
$begingroup$
@Surb i fix the differentials, anyway it was obvious from the proof that it's $dy$ because I consider $f(x)$ as constant under the integral. And monotone convergence theorem is for increasing sequences and that sequence is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:37
1
$begingroup$
Yes, I was thinking about Dominated theorem convergence. Anyway, it's almost straightforward ;-)
$endgroup$
– Surb
Jan 14 at 17:40
$begingroup$
@Surb I think it should work as well. I already prove it with your assumption. It will look weird tho. Lol. I proved that $lim_{ntoinfty} int g_n(x) = 0$ then the $limsup = 0$ as well.
$endgroup$
– Richard Clare
Jan 14 at 17:46
add a comment |
$begingroup$
I must show that
Let $f$ be Lebesgueable integrable, then
$$lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy = 0$$
for almost every $x in mathbb{R}$.
Tentative proof:
$$|f(x) - f(y)| geq 0 Rightarrow lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy geq 0$$
We must show that
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy leq 0$$
I will show that for any $epsilon >0$
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy < epsilon$$
If $f(x) = 0$, By absolute continuity of integration, for any $epsilon >0$, $exists delta >0$ such that if
$$ m({|x-y| < h}) < delta Rightarrow int_{|x -y| < h}|f(y)|dy <frac{epsilon}{2} $$
Now $$ lim_{hto 0}sup m({|x-y| < h}) leq m({|x-y| < h}) < delta$$
then
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy < frac{epsilon}{2} < epsilon $$
Now for $f(x) neq 0$,
$$int_{|x -y| < h}|f(x) - f(y)|dy leq int_{|x -y| < h}|f(x)|dy + int_{|x -y| < h}|f(y)|dy $$
$$= |f(x)|m({|x -y| < h}) + int_{|x -y| < h}|f(y)|dy $$
and $|f(x)| < infty$ for allmost all $x in mathbb{R}$, since $f$ is Lebesgue integrable.
Let $delta^{*} < min {delta,frac{epsilon}{2|f(x)|}}$, now by absolute continuiuty of integration, for $epsilon > 0$,
$$m({|x -y| < h}) < delta^{*} < delta Rightarrow int_{|x -y| < h}|f(x) - f(y)|dy leq |f(x)|m({|x -y| < h}) + int_{|x -y| < h}|f(y)|dy < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$$
For almost every $x in mathbb{R}$. Hence,
$$ lim_{hto 0}sup m({|x-y| < h}) leq m({|x-y| < h}) < delta^{*} < delta Rightarrow lim_{hto 0}supint_{|x -y| < h}|f(x) - f(y)|dy < epsilon $$
Therefore,
$$lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy = 0$$
for almost every $x in mathbb{R}$.
lebesgue-integral lebesgue-measure limsup-and-liminf absolute-continuity
asked Jan 14 at 17:09
Richard ClareRichard Clare
1,076314
$endgroup$
I must show that
Let $f$ be Lebesgueable integrable, then
$$lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy = 0$$
for almost every $x in mathbb{R}$.
Tentative proof:
$$|f(x) - f(y)| geq 0 Rightarrow lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy geq 0$$
We must show that
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy leq 0$$
I will show that for any $epsilon >0$
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy < epsilon$$
If $f(x) = 0$, By absolute continuity of integration, for any $epsilon >0$, $exists delta >0$ such that if
$$ m({|x-y| < h}) < delta Rightarrow int_{|x -y| < h}|f(y)|dy <frac{epsilon}{2} $$
Now $$ lim_{hto 0}sup m({|x-y| < h}) leq m({|x-y| < h}) < delta$$
then
$$ lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy < frac{epsilon}{2} < epsilon $$
Now for $f(x) neq 0$,
$$int_{|x -y| < h}|f(x) - f(y)|dy leq int_{|x -y| < h}|f(x)|dy + int_{|x -y| < h}|f(y)|dy $$
$$= |f(x)|m({|x -y| < h}) + int_{|x -y| < h}|f(y)|dy $$
and $|f(x)| < infty$ for allmost all $x in mathbb{R}$, since $f$ is Lebesgue integrable.
Let $delta^{*} < min {delta,frac{epsilon}{2|f(x)|}}$, now by absolute continuiuty of integration, for $epsilon > 0$,
$$m({|x -y| < h}) < delta^{*} < delta Rightarrow int_{|x -y| < h}|f(x) - f(y)|dy leq |f(x)|m({|x -y| < h}) + int_{|x -y| < h}|f(y)|dy < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$$
For almost every $x in mathbb{R}$. Hence,
$$ lim_{hto 0}sup m({|x-y| < h}) leq m({|x-y| < h}) < delta^{*} < delta Rightarrow lim_{hto 0}supint_{|x -y| < h}|f(x) - f(y)|dy < epsilon $$
Therefore,
$$lim_{hto 0}sup int_{|x -y| < h}|f(x) - f(y)|dy = 0$$
for almost every $x in mathbb{R}$.
lebesgue-integral lebesgue-measure limsup-and-liminf absolute-continuity
lebesgue-integral lebesgue-measure limsup-and-liminf absolute-continuity
asked Jan 14 at 17:09
Richard ClareRichard Clare
1,076314
asked Jan 14 at 17:09
Richard ClareRichard Clare
1,076314
edited Jan 14 at 17:35
Richard Clare
asked Jan 14 at 17:09
Richard ClareRichard Clare
1,076314
asked Jan 14 at 17:09
Richard ClareRichard Clare
1,076314
asked Jan 14 at 17:09
Richard ClareRichard Clare
1,076314
1,076314
1
$begingroup$
Sorry but is a bit complicate to read your proof (too long ^^). btw, it's not clear if you integrate wrt $x$, $y$ or both. Anyway, for me it's just monotone convergence with $g_n(x,y)=|f(x)-f(y)|boldsymbol 1_{{|x-y|<frac{1}{n}}}$. Should work, no ?
$endgroup$
– Surb
Jan 14 at 17:30
$begingroup$
its with respect to $y$. And I see what you mean, but that set is not increasing $|x-y| < 1/n$ is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:34
$begingroup$
@Surb i fix the differentials, anyway it was obvious from the proof that it's $dy$ because I consider $f(x)$ as constant under the integral. And monotone convergence theorem is for increasing sequences and that sequence is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:37
1
$begingroup$
Yes, I was thinking about Dominated theorem convergence. Anyway, it's almost straightforward ;-)
$endgroup$
– Surb
Jan 14 at 17:40
$begingroup$
@Surb I think it should work as well. I already prove it with your assumption. It will look weird tho. Lol. I proved that $lim_{ntoinfty} int g_n(x) = 0$ then the $limsup = 0$ as well.
$endgroup$
– Richard Clare
Jan 14 at 17:46
add a comment |
1
$begingroup$
Sorry but is a bit complicate to read your proof (too long ^^). btw, it's not clear if you integrate wrt $x$, $y$ or both. Anyway, for me it's just monotone convergence with $g_n(x,y)=|f(x)-f(y)|boldsymbol 1_{{|x-y|<frac{1}{n}}}$. Should work, no ?
$endgroup$
– Surb
Jan 14 at 17:30
$begingroup$
its with respect to $y$. And I see what you mean, but that set is not increasing $|x-y| < 1/n$ is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:34
$begingroup$
@Surb i fix the differentials, anyway it was obvious from the proof that it's $dy$ because I consider $f(x)$ as constant under the integral. And monotone convergence theorem is for increasing sequences and that sequence is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:37
1
$begingroup$
Yes, I was thinking about Dominated theorem convergence. Anyway, it's almost straightforward ;-)
$endgroup$
– Surb
Jan 14 at 17:40
$begingroup$
@Surb I think it should work as well. I already prove it with your assumption. It will look weird tho. Lol. I proved that $lim_{ntoinfty} int g_n(x) = 0$ then the $limsup = 0$ as well.
$endgroup$
– Richard Clare
Jan 14 at 17:46
1
1
$begingroup$
Sorry but is a bit complicate to read your proof (too long ^^). btw, it's not clear if you integrate wrt $x$, $y$ or both. Anyway, for me it's just monotone convergence with $g_n(x,y)=|f(x)-f(y)|boldsymbol 1_{{|x-y|<frac{1}{n}}}$. Should work, no ?
$endgroup$
– Surb
Jan 14 at 17:30
$begingroup$
Sorry but is a bit complicate to read your proof (too long ^^). btw, it's not clear if you integrate wrt $x$, $y$ or both. Anyway, for me it's just monotone convergence with $g_n(x,y)=|f(x)-f(y)|boldsymbol 1_{{|x-y|<frac{1}{n}}}$. Should work, no ?
$endgroup$
– Surb
Jan 14 at 17:30
$begingroup$
its with respect to $y$. And I see what you mean, but that set is not increasing $|x-y| < 1/n$ is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:34
$begingroup$
its with respect to $y$. And I see what you mean, but that set is not increasing $|x-y| < 1/n$ is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:34
$begingroup$
@Surb i fix the differentials, anyway it was obvious from the proof that it's $dy$ because I consider $f(x)$ as constant under the integral. And monotone convergence theorem is for increasing sequences and that sequence is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:37
$begingroup$
@Surb i fix the differentials, anyway it was obvious from the proof that it's $dy$ because I consider $f(x)$ as constant under the integral. And monotone convergence theorem is for increasing sequences and that sequence is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:37
1
1
$begingroup$
Yes, I was thinking about Dominated theorem convergence. Anyway, it's almost straightforward ;-)
$endgroup$
– Surb
Jan 14 at 17:40
$begingroup$
Yes, I was thinking about Dominated theorem convergence. Anyway, it's almost straightforward ;-)
$endgroup$
– Surb
Jan 14 at 17:40
$begingroup$
@Surb I think it should work as well. I already prove it with your assumption. It will look weird tho. Lol. I proved that $lim_{ntoinfty} int g_n(x) = 0$ then the $limsup = 0$ as well.
$endgroup$
– Richard Clare
Jan 14 at 17:46
$begingroup$
@Surb I think it should work as well. I already prove it with your assumption. It will look weird tho. Lol. I proved that $lim_{ntoinfty} int g_n(x) = 0$ then the $limsup = 0$ as well.
$endgroup$
– Richard Clare
Jan 14 at 17:46
add a comment |
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$begingroup$
Sorry but is a bit complicate to read your proof (too long ^^). btw, it's not clear if you integrate wrt $x$, $y$ or both. Anyway, for me it's just monotone convergence with $g_n(x,y)=|f(x)-f(y)|boldsymbol 1_{{|x-y|<frac{1}{n}}}$. Should work, no ?
$endgroup$
– Surb
Jan 14 at 17:30
$begingroup$
its with respect to $y$. And I see what you mean, but that set is not increasing $|x-y| < 1/n$ is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:34
$begingroup$
@Surb i fix the differentials, anyway it was obvious from the proof that it's $dy$ because I consider $f(x)$ as constant under the integral. And monotone convergence theorem is for increasing sequences and that sequence is decreasing.
$endgroup$
– Richard Clare
Jan 14 at 17:37
1
$begingroup$
Yes, I was thinking about Dominated theorem convergence. Anyway, it's almost straightforward ;-)
$endgroup$
– Surb
Jan 14 at 17:40
$begingroup$
@Surb I think it should work as well. I already prove it with your assumption. It will look weird tho. Lol. I proved that $lim_{ntoinfty} int g_n(x) = 0$ then the $limsup = 0$ as well.
$endgroup$
– Richard Clare
Jan 14 at 17:46