Can the supremum of continuous functions be discontinuous at every point of an interval?

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Pether Luthy gave an example of a sequence of continuous real valued functions whose supremum was discontinuous on a set of positive measure. But does it exist a sequence of continuous real valued functions $f_n:mathbbRtomathbbR$ such that $f(x) = sup_n in mathbbN f_n(x)$ is a discontinuous function at every point of a subinterval of $mathbbR$ ?



If such a sequence does not exist, how is it possible to prove it?






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  • Are you asking for a sequence of continuous real valued functions $f_n$ such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$?
    – Stanley Yao Xiao
    Aug 31 at 23:51






  • 1




    When you mention Peter Luthy's example, you mean the answer to this question, right: Can the supremum of continuous functions be discontinuous on a set of positive measure?
    – Martin Sleziak
    Aug 31 at 23:51










  • Yes, I am asking for a sequence of continuous real valued functions such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$
    – Angelo
    Aug 31 at 23:54










  • Yes, Martin, I mean the answer to that question.
    – Angelo
    Aug 31 at 23:55










  • @StanleyYaoXiao The OP wants the function to be discontinuous at every point in the interval.
    – Noah Schweber
    Aug 31 at 23:58














up vote
6
down vote

favorite












Pether Luthy gave an example of a sequence of continuous real valued functions whose supremum was discontinuous on a set of positive measure. But does it exist a sequence of continuous real valued functions $f_n:mathbbRtomathbbR$ such that $f(x) = sup_n in mathbbN f_n(x)$ is a discontinuous function at every point of a subinterval of $mathbbR$ ?



If such a sequence does not exist, how is it possible to prove it?






real-analysis







share|cite|improve this question























  • Are you asking for a sequence of continuous real valued functions $f_n$ such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$?
    – Stanley Yao Xiao
    Aug 31 at 23:51






  • 1




    When you mention Peter Luthy's example, you mean the answer to this question, right: Can the supremum of continuous functions be discontinuous on a set of positive measure?
    – Martin Sleziak
    Aug 31 at 23:51










  • Yes, I am asking for a sequence of continuous real valued functions such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$
    – Angelo
    Aug 31 at 23:54










  • Yes, Martin, I mean the answer to that question.
    – Angelo
    Aug 31 at 23:55










  • @StanleyYaoXiao The OP wants the function to be discontinuous at every point in the interval.
    – Noah Schweber
    Aug 31 at 23:58












up vote
6
down vote

favorite









up vote
6
down vote

favorite











Pether Luthy gave an example of a sequence of continuous real valued functions whose supremum was discontinuous on a set of positive measure. But does it exist a sequence of continuous real valued functions $f_n:mathbbRtomathbbR$ such that $f(x) = sup_n in mathbbN f_n(x)$ is a discontinuous function at every point of a subinterval of $mathbbR$ ?



If such a sequence does not exist, how is it possible to prove it?






real-analysis







share|cite|improve this question















Pether Luthy gave an example of a sequence of continuous real valued functions whose supremum was discontinuous on a set of positive measure. But does it exist a sequence of continuous real valued functions $f_n:mathbbRtomathbbR$ such that $f(x) = sup_n in mathbbN f_n(x)$ is a discontinuous function at every point of a subinterval of $mathbbR$ ?



If such a sequence does not exist, how is it possible to prove it?






real-analysis




real-analysis






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share|cite|improve this question













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edited Sep 1 at 0:13

























asked Aug 31 at 23:45









Angelo

312




312











  • Are you asking for a sequence of continuous real valued functions $f_n$ such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$?
    – Stanley Yao Xiao
    Aug 31 at 23:51






  • 1




    When you mention Peter Luthy's example, you mean the answer to this question, right: Can the supremum of continuous functions be discontinuous on a set of positive measure?
    – Martin Sleziak
    Aug 31 at 23:51










  • Yes, I am asking for a sequence of continuous real valued functions such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$
    – Angelo
    Aug 31 at 23:54










  • Yes, Martin, I mean the answer to that question.
    – Angelo
    Aug 31 at 23:55










  • @StanleyYaoXiao The OP wants the function to be discontinuous at every point in the interval.
    – Noah Schweber
    Aug 31 at 23:58
















  • Are you asking for a sequence of continuous real valued functions $f_n$ such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$?
    – Stanley Yao Xiao
    Aug 31 at 23:51






  • 1




    When you mention Peter Luthy's example, you mean the answer to this question, right: Can the supremum of continuous functions be discontinuous on a set of positive measure?
    – Martin Sleziak
    Aug 31 at 23:51










  • Yes, I am asking for a sequence of continuous real valued functions such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$
    – Angelo
    Aug 31 at 23:54










  • Yes, Martin, I mean the answer to that question.
    – Angelo
    Aug 31 at 23:55










  • @StanleyYaoXiao The OP wants the function to be discontinuous at every point in the interval.
    – Noah Schweber
    Aug 31 at 23:58















Are you asking for a sequence of continuous real valued functions $f_n$ such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$?
– Stanley Yao Xiao
Aug 31 at 23:51




Are you asking for a sequence of continuous real valued functions $f_n$ such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$?
– Stanley Yao Xiao
Aug 31 at 23:51




1




1




When you mention Peter Luthy's example, you mean the answer to this question, right: Can the supremum of continuous functions be discontinuous on a set of positive measure?
– Martin Sleziak
Aug 31 at 23:51




When you mention Peter Luthy's example, you mean the answer to this question, right: Can the supremum of continuous functions be discontinuous on a set of positive measure?
– Martin Sleziak
Aug 31 at 23:51












Yes, I am asking for a sequence of continuous real valued functions such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$
– Angelo
Aug 31 at 23:54




Yes, I am asking for a sequence of continuous real valued functions such that $f(x) = sup_n in mathbbN f_n(x)$ is discontinuous on a subinterval of $mathbbR$
– Angelo
Aug 31 at 23:54












Yes, Martin, I mean the answer to that question.
– Angelo
Aug 31 at 23:55




Yes, Martin, I mean the answer to that question.
– Angelo
Aug 31 at 23:55












@StanleyYaoXiao The OP wants the function to be discontinuous at every point in the interval.
– Noah Schweber
Aug 31 at 23:58




@StanleyYaoXiao The OP wants the function to be discontinuous at every point in the interval.
– Noah Schweber
Aug 31 at 23:58










1 Answer
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Since the function $f$ is supremum of a set of continuous functions, it is lower-semicontinuous.1



Every lower semicontinuous function belongs to the first Baire class.2



If $fcolon mathbb Rtomathbb R$ is of the first Baire class, then the set $D_f$ of the points of discontinuity is a meager set.3



In particular, $D_f$ cannot be an interval.



1Theorem 10.3 in van Rooij-Schikhof: A Second Course on Real Functions. Mathematics Stack Exchange: To show that the supremum of any collection of lower semicontinuous functions is lower semicontinuous or Show that the supremum of a collection of lower semicontinuous function is lower semicontinuous.



2Theorem 10.6 and Exercise 11.E in van Rooij-Schikhof; Show that lower semicontinuous function is the supremum of an increasing sequence of continuous functions on Mathematics Stack Exchange



3Theorem 11.4 in van Rooij-Schikhof; MathOverflow: Points of continuity of Baire class one functions






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  • 1




    One can bypass the ``lsc implies Baire-1'' step by noting directly that Angelo's $f$ is Baire-1, since one has for $g_n = supf_1,dots,f_n$ that $f=lim g_n$ pointwise.
    – Dirk Werner
    Sep 4 at 18:54










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
23
down vote













Since the function $f$ is supremum of a set of continuous functions, it is lower-semicontinuous.1



Every lower semicontinuous function belongs to the first Baire class.2



If $fcolon mathbb Rtomathbb R$ is of the first Baire class, then the set $D_f$ of the points of discontinuity is a meager set.3



In particular, $D_f$ cannot be an interval.



1Theorem 10.3 in van Rooij-Schikhof: A Second Course on Real Functions. Mathematics Stack Exchange: To show that the supremum of any collection of lower semicontinuous functions is lower semicontinuous or Show that the supremum of a collection of lower semicontinuous function is lower semicontinuous.



2Theorem 10.6 and Exercise 11.E in van Rooij-Schikhof; Show that lower semicontinuous function is the supremum of an increasing sequence of continuous functions on Mathematics Stack Exchange



3Theorem 11.4 in van Rooij-Schikhof; MathOverflow: Points of continuity of Baire class one functions






share|cite|improve this answer


















  • 1




    One can bypass the ``lsc implies Baire-1'' step by noting directly that Angelo's $f$ is Baire-1, since one has for $g_n = supf_1,dots,f_n$ that $f=lim g_n$ pointwise.
    – Dirk Werner
    Sep 4 at 18:54














up vote
23
down vote













Since the function $f$ is supremum of a set of continuous functions, it is lower-semicontinuous.1



Every lower semicontinuous function belongs to the first Baire class.2



If $fcolon mathbb Rtomathbb R$ is of the first Baire class, then the set $D_f$ of the points of discontinuity is a meager set.3



In particular, $D_f$ cannot be an interval.



1Theorem 10.3 in van Rooij-Schikhof: A Second Course on Real Functions. Mathematics Stack Exchange: To show that the supremum of any collection of lower semicontinuous functions is lower semicontinuous or Show that the supremum of a collection of lower semicontinuous function is lower semicontinuous.



2Theorem 10.6 and Exercise 11.E in van Rooij-Schikhof; Show that lower semicontinuous function is the supremum of an increasing sequence of continuous functions on Mathematics Stack Exchange



3Theorem 11.4 in van Rooij-Schikhof; MathOverflow: Points of continuity of Baire class one functions






share|cite|improve this answer


















  • 1




    One can bypass the ``lsc implies Baire-1'' step by noting directly that Angelo's $f$ is Baire-1, since one has for $g_n = supf_1,dots,f_n$ that $f=lim g_n$ pointwise.
    – Dirk Werner
    Sep 4 at 18:54












up vote
23
down vote










up vote
23
down vote









Since the function $f$ is supremum of a set of continuous functions, it is lower-semicontinuous.1



Every lower semicontinuous function belongs to the first Baire class.2



If $fcolon mathbb Rtomathbb R$ is of the first Baire class, then the set $D_f$ of the points of discontinuity is a meager set.3



In particular, $D_f$ cannot be an interval.



1Theorem 10.3 in van Rooij-Schikhof: A Second Course on Real Functions. Mathematics Stack Exchange: To show that the supremum of any collection of lower semicontinuous functions is lower semicontinuous or Show that the supremum of a collection of lower semicontinuous function is lower semicontinuous.



2Theorem 10.6 and Exercise 11.E in van Rooij-Schikhof; Show that lower semicontinuous function is the supremum of an increasing sequence of continuous functions on Mathematics Stack Exchange



3Theorem 11.4 in van Rooij-Schikhof; MathOverflow: Points of continuity of Baire class one functions






share|cite|improve this answer














Since the function $f$ is supremum of a set of continuous functions, it is lower-semicontinuous.1



Every lower semicontinuous function belongs to the first Baire class.2



If $fcolon mathbb Rtomathbb R$ is of the first Baire class, then the set $D_f$ of the points of discontinuity is a meager set.3



In particular, $D_f$ cannot be an interval.



1Theorem 10.3 in van Rooij-Schikhof: A Second Course on Real Functions. Mathematics Stack Exchange: To show that the supremum of any collection of lower semicontinuous functions is lower semicontinuous or Show that the supremum of a collection of lower semicontinuous function is lower semicontinuous.



2Theorem 10.6 and Exercise 11.E in van Rooij-Schikhof; Show that lower semicontinuous function is the supremum of an increasing sequence of continuous functions on Mathematics Stack Exchange



3Theorem 11.4 in van Rooij-Schikhof; MathOverflow: Points of continuity of Baire class one functions







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 1 at 7:11

























answered Sep 1 at 0:31









Martin Sleziak

2,74432028




2,74432028







  • 1




    One can bypass the ``lsc implies Baire-1'' step by noting directly that Angelo's $f$ is Baire-1, since one has for $g_n = supf_1,dots,f_n$ that $f=lim g_n$ pointwise.
    – Dirk Werner
    Sep 4 at 18:54












  • 1




    One can bypass the ``lsc implies Baire-1'' step by noting directly that Angelo's $f$ is Baire-1, since one has for $g_n = supf_1,dots,f_n$ that $f=lim g_n$ pointwise.
    – Dirk Werner
    Sep 4 at 18:54







1




1




One can bypass the ``lsc implies Baire-1'' step by noting directly that Angelo's $f$ is Baire-1, since one has for $g_n = supf_1,dots,f_n$ that $f=lim g_n$ pointwise.
– Dirk Werner
Sep 4 at 18:54




One can bypass the ``lsc implies Baire-1'' step by noting directly that Angelo's $f$ is Baire-1, since one has for $g_n = supf_1,dots,f_n$ that $f=lim g_n$ pointwise.
– Dirk Werner
Sep 4 at 18:54

















 

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