Riemann integrable function implies discontinuous on a Borel set?
$begingroup$
In this post I explain the proof, by which I prove that if $f $ is Riemann integrable function on [a,b] $implies f$ discontinuous on a Borel set.
I know, that in theory, this is wrong.
Question : Where have I made the wrong argument, and why?
One approach to proving a Riemann function($f$ on $[a,b ]$) is Lebesgue integrable is by using a nested partition, $P_0 supseteq P_1 supseteq P_2 ...$ with the $lim_{nto infty} |P_n|=0$. Afterwards the associated step functions are considered, $h_n: [a,b] rightarrow mathbb{R}, xrightarrow m_i, m_i=inf(f,[e_i,e_{i+1}]), x in [e_i,e_{i+1}), {e_i,e_{i+1}} subset P_n$, and $b rightarrow m_j=inf(f,[e_j,e_{j+1}]), b in [e_j,e_{j+1}], {e_j,e_{j+1}} subset P_n$, obviously here $e_{j+1}=b$. Similarly $g_n$ is constructed only instead of $m_i$ the $M_i=sup(f,[e_i,e_{i+1}])$ is considered.
Then the following statements are known to be true:
$A={ x | f(x)=lim h_n(x)=lim g_n(x) }$ is Borel measurable. $B=Xsetminus A$
$f$ is continuous in $[a,b] setminus (B cup Delta)$, where $Delta=cup P_n$ which is also countable, therefore any subset of $Delta$ is Borel.
And then I prove $f$ is discontinuous on $Bsetminus Delta$. By, if $x in Bsetminus Delta$, then exist a nested sequence of $[a_1, b_1) supseteq [a_2, b_2) ...$ from the corresponding partitions, whose limit of diameters goes to $0$, and $x$ is in the interior of each half-open interval. Since in this case $lim h_n(x)neq lim g_n(x)$, the oscillation of $f$ at $x$ won't be zero, therefore $f$ discontinuous at $x$. Hence $f$, is discontinuous on $Bsetminus Delta$, which is Borel, and also can "only"( it is not discontinous on $[a,b] setminus (B cup Delta)$) be discontinuous on a subset from $Delta$, which is also Borel.
And therefore, using (2), I conclude that $f$ is Riemann integrable implies $f$ discontinuous on a Borel set.
continuity lebesgue-integral lebesgue-measure riemann-integration borel-sets
edited Dec 31 '18 at 11:01
Davide Giraudo
127k17154268
asked Dec 18 '18 at 0:20
fseppfsepp
163
$endgroup$
add a comment |
$begingroup$
In this post I explain the proof, by which I prove that if $f $ is Riemann integrable function on [a,b] $implies f$ discontinuous on a Borel set.
I know, that in theory, this is wrong.
Question : Where have I made the wrong argument, and why?
One approach to proving a Riemann function($f$ on $[a,b ]$) is Lebesgue integrable is by using a nested partition, $P_0 supseteq P_1 supseteq P_2 ...$ with the $lim_{nto infty} |P_n|=0$. Afterwards the associated step functions are considered, $h_n: [a,b] rightarrow mathbb{R}, xrightarrow m_i, m_i=inf(f,[e_i,e_{i+1}]), x in [e_i,e_{i+1}), {e_i,e_{i+1}} subset P_n$, and $b rightarrow m_j=inf(f,[e_j,e_{j+1}]), b in [e_j,e_{j+1}], {e_j,e_{j+1}} subset P_n$, obviously here $e_{j+1}=b$. Similarly $g_n$ is constructed only instead of $m_i$ the $M_i=sup(f,[e_i,e_{i+1}])$ is considered.
Then the following statements are known to be true:
$A={ x | f(x)=lim h_n(x)=lim g_n(x) }$ is Borel measurable. $B=Xsetminus A$
$f$ is continuous in $[a,b] setminus (B cup Delta)$, where $Delta=cup P_n$ which is also countable, therefore any subset of $Delta$ is Borel.
And then I prove $f$ is discontinuous on $Bsetminus Delta$. By, if $x in Bsetminus Delta$, then exist a nested sequence of $[a_1, b_1) supseteq [a_2, b_2) ...$ from the corresponding partitions, whose limit of diameters goes to $0$, and $x$ is in the interior of each half-open interval. Since in this case $lim h_n(x)neq lim g_n(x)$, the oscillation of $f$ at $x$ won't be zero, therefore $f$ discontinuous at $x$. Hence $f$, is discontinuous on $Bsetminus Delta$, which is Borel, and also can "only"( it is not discontinous on $[a,b] setminus (B cup Delta)$) be discontinuous on a subset from $Delta$, which is also Borel.
And therefore, using (2), I conclude that $f$ is Riemann integrable implies $f$ discontinuous on a Borel set.
continuity lebesgue-integral lebesgue-measure riemann-integration borel-sets
edited Dec 31 '18 at 11:01
Davide Giraudo
127k17154268
asked Dec 18 '18 at 0:20
fseppfsepp
163
$endgroup$
2
$begingroup$
What does "$f$ is discontinuous on a Borel set$ mean? Does it mean "the set of all points at which $f$ is discontinuous is a Borel set"? That is true for every function $f:mathbb Rtomathbb R$. In fact the set of points of discontinuity of an arbitrary real function is an $F_sigma$ set. If $f$ is Riemann integrable, we can say more: the set of points of discontinuity has Lebesgue measure zero.
$endgroup$
– bof
Dec 31 '18 at 11:05
$begingroup$
Yes, and you are right. I thought( at least I admitted ) that there exist functions that were discontinuous on [a,b ] on a Lebesgue null set, but which is not Borel. But such functions do not exist.
$endgroup$
– fsepp
Jan 3 at 1:34
add a comment |
$begingroup$
In this post I explain the proof, by which I prove that if $f $ is Riemann integrable function on [a,b] $implies f$ discontinuous on a Borel set.
I know, that in theory, this is wrong.
Question : Where have I made the wrong argument, and why?
One approach to proving a Riemann function($f$ on $[a,b ]$) is Lebesgue integrable is by using a nested partition, $P_0 supseteq P_1 supseteq P_2 ...$ with the $lim_{nto infty} |P_n|=0$. Afterwards the associated step functions are considered, $h_n: [a,b] rightarrow mathbb{R}, xrightarrow m_i, m_i=inf(f,[e_i,e_{i+1}]), x in [e_i,e_{i+1}), {e_i,e_{i+1}} subset P_n$, and $b rightarrow m_j=inf(f,[e_j,e_{j+1}]), b in [e_j,e_{j+1}], {e_j,e_{j+1}} subset P_n$, obviously here $e_{j+1}=b$. Similarly $g_n$ is constructed only instead of $m_i$ the $M_i=sup(f,[e_i,e_{i+1}])$ is considered.
Then the following statements are known to be true:
$A={ x | f(x)=lim h_n(x)=lim g_n(x) }$ is Borel measurable. $B=Xsetminus A$
$f$ is continuous in $[a,b] setminus (B cup Delta)$, where $Delta=cup P_n$ which is also countable, therefore any subset of $Delta$ is Borel.
And then I prove $f$ is discontinuous on $Bsetminus Delta$. By, if $x in Bsetminus Delta$, then exist a nested sequence of $[a_1, b_1) supseteq [a_2, b_2) ...$ from the corresponding partitions, whose limit of diameters goes to $0$, and $x$ is in the interior of each half-open interval. Since in this case $lim h_n(x)neq lim g_n(x)$, the oscillation of $f$ at $x$ won't be zero, therefore $f$ discontinuous at $x$. Hence $f$, is discontinuous on $Bsetminus Delta$, which is Borel, and also can "only"( it is not discontinous on $[a,b] setminus (B cup Delta)$) be discontinuous on a subset from $Delta$, which is also Borel.
And therefore, using (2), I conclude that $f$ is Riemann integrable implies $f$ discontinuous on a Borel set.
continuity lebesgue-integral lebesgue-measure riemann-integration borel-sets
edited Dec 31 '18 at 11:01
Davide Giraudo
127k17154268
asked Dec 18 '18 at 0:20
fseppfsepp
163
$endgroup$
In this post I explain the proof, by which I prove that if $f $ is Riemann integrable function on [a,b] $implies f$ discontinuous on a Borel set.
I know, that in theory, this is wrong.
Question : Where have I made the wrong argument, and why?
One approach to proving a Riemann function($f$ on $[a,b ]$) is Lebesgue integrable is by using a nested partition, $P_0 supseteq P_1 supseteq P_2 ...$ with the $lim_{nto infty} |P_n|=0$. Afterwards the associated step functions are considered, $h_n: [a,b] rightarrow mathbb{R}, xrightarrow m_i, m_i=inf(f,[e_i,e_{i+1}]), x in [e_i,e_{i+1}), {e_i,e_{i+1}} subset P_n$, and $b rightarrow m_j=inf(f,[e_j,e_{j+1}]), b in [e_j,e_{j+1}], {e_j,e_{j+1}} subset P_n$, obviously here $e_{j+1}=b$. Similarly $g_n$ is constructed only instead of $m_i$ the $M_i=sup(f,[e_i,e_{i+1}])$ is considered.
Then the following statements are known to be true:
$A={ x | f(x)=lim h_n(x)=lim g_n(x) }$ is Borel measurable. $B=Xsetminus A$
$f$ is continuous in $[a,b] setminus (B cup Delta)$, where $Delta=cup P_n$ which is also countable, therefore any subset of $Delta$ is Borel.
And then I prove $f$ is discontinuous on $Bsetminus Delta$. By, if $x in Bsetminus Delta$, then exist a nested sequence of $[a_1, b_1) supseteq [a_2, b_2) ...$ from the corresponding partitions, whose limit of diameters goes to $0$, and $x$ is in the interior of each half-open interval. Since in this case $lim h_n(x)neq lim g_n(x)$, the oscillation of $f$ at $x$ won't be zero, therefore $f$ discontinuous at $x$. Hence $f$, is discontinuous on $Bsetminus Delta$, which is Borel, and also can "only"( it is not discontinous on $[a,b] setminus (B cup Delta)$) be discontinuous on a subset from $Delta$, which is also Borel.
And therefore, using (2), I conclude that $f$ is Riemann integrable implies $f$ discontinuous on a Borel set.
continuity lebesgue-integral lebesgue-measure riemann-integration borel-sets
continuity lebesgue-integral lebesgue-measure riemann-integration borel-sets
edited Dec 31 '18 at 11:01
Davide Giraudo
127k17154268
asked Dec 18 '18 at 0:20
fseppfsepp
163
edited Dec 31 '18 at 11:01
Davide Giraudo
127k17154268
asked Dec 18 '18 at 0:20
fseppfsepp
163
edited Dec 31 '18 at 11:01
Davide Giraudo
127k17154268
edited Dec 31 '18 at 11:01
Davide Giraudo
127k17154268
edited Dec 31 '18 at 11:01
Davide Giraudo
127k17154268
127k17154268
asked Dec 18 '18 at 0:20
fseppfsepp
163
asked Dec 18 '18 at 0:20
fseppfsepp
163
asked Dec 18 '18 at 0:20
fseppfsepp
163
163
2
$begingroup$
What does "$f$ is discontinuous on a Borel set$ mean? Does it mean "the set of all points at which $f$ is discontinuous is a Borel set"? That is true for every function $f:mathbb Rtomathbb R$. In fact the set of points of discontinuity of an arbitrary real function is an $F_sigma$ set. If $f$ is Riemann integrable, we can say more: the set of points of discontinuity has Lebesgue measure zero.
$endgroup$
– bof
Dec 31 '18 at 11:05
$begingroup$
Yes, and you are right. I thought( at least I admitted ) that there exist functions that were discontinuous on [a,b ] on a Lebesgue null set, but which is not Borel. But such functions do not exist.
$endgroup$
– fsepp
Jan 3 at 1:34
add a comment |
2
$begingroup$
What does "$f$ is discontinuous on a Borel set$ mean? Does it mean "the set of all points at which $f$ is discontinuous is a Borel set"? That is true for every function $f:mathbb Rtomathbb R$. In fact the set of points of discontinuity of an arbitrary real function is an $F_sigma$ set. If $f$ is Riemann integrable, we can say more: the set of points of discontinuity has Lebesgue measure zero.
$endgroup$
– bof
Dec 31 '18 at 11:05
$begingroup$
Yes, and you are right. I thought( at least I admitted ) that there exist functions that were discontinuous on [a,b ] on a Lebesgue null set, but which is not Borel. But such functions do not exist.
$endgroup$
– fsepp
Jan 3 at 1:34
2
2
$begingroup$
What does "$f$ is discontinuous on a Borel set$ mean? Does it mean "the set of all points at which $f$ is discontinuous is a Borel set"? That is true for every function $f:mathbb Rtomathbb R$. In fact the set of points of discontinuity of an arbitrary real function is an $F_sigma$ set. If $f$ is Riemann integrable, we can say more: the set of points of discontinuity has Lebesgue measure zero.
$endgroup$
– bof
Dec 31 '18 at 11:05
$begingroup$
What does "$f$ is discontinuous on a Borel set$ mean? Does it mean "the set of all points at which $f$ is discontinuous is a Borel set"? That is true for every function $f:mathbb Rtomathbb R$. In fact the set of points of discontinuity of an arbitrary real function is an $F_sigma$ set. If $f$ is Riemann integrable, we can say more: the set of points of discontinuity has Lebesgue measure zero.
$endgroup$
– bof
Dec 31 '18 at 11:05
$begingroup$
Yes, and you are right. I thought( at least I admitted ) that there exist functions that were discontinuous on [a,b ] on a Lebesgue null set, but which is not Borel. But such functions do not exist.
$endgroup$
– fsepp
Jan 3 at 1:34
$begingroup$
Yes, and you are right. I thought( at least I admitted ) that there exist functions that were discontinuous on [a,b ] on a Lebesgue null set, but which is not Borel. But such functions do not exist.
$endgroup$
– fsepp
Jan 3 at 1:34
add a comment |
0
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$begingroup$
What does "$f$ is discontinuous on a Borel set$ mean? Does it mean "the set of all points at which $f$ is discontinuous is a Borel set"? That is true for every function $f:mathbb Rtomathbb R$. In fact the set of points of discontinuity of an arbitrary real function is an $F_sigma$ set. If $f$ is Riemann integrable, we can say more: the set of points of discontinuity has Lebesgue measure zero.
$endgroup$
– bof
Dec 31 '18 at 11:05
$begingroup$
Yes, and you are right. I thought( at least I admitted ) that there exist functions that were discontinuous on [a,b ] on a Lebesgue null set, but which is not Borel. But such functions do not exist.
$endgroup$
– fsepp
Jan 3 at 1:34