Mixing
What is the probability of a function being Riemann Integrable?
$begingroup$
Suppose we are given a set $F ={f| f:mathbb{R} to mathbb{R}}$, i.e. an arbitrary set (not sure about its countablility) of real-valued functions. What is the probability, that a function $f_k$, chosen at random from the set is Riemann Integrable?
Note: I came up with this problem entirely on my own. So, if there are any "loose-statements", please correct it. I don't have enough knowledge to even approach this. There might be some duplicates out there, which I am not sure of. Is this even a valid question?
Thank you.
real-analysis probability functions soft-question riemann-integration
asked Jan 26 at 15:24
Subhasis BiswasSubhasis Biswas
511411
$endgroup$
|
show 6 more comments
$begingroup$
Suppose we are given a set $F ={f| f:mathbb{R} to mathbb{R}}$, i.e. an arbitrary set (not sure about its countablility) of real-valued functions. What is the probability, that a function $f_k$, chosen at random from the set is Riemann Integrable?
Note: I came up with this problem entirely on my own. So, if there are any "loose-statements", please correct it. I don't have enough knowledge to even approach this. There might be some duplicates out there, which I am not sure of. Is this even a valid question?
Thank you.
real-analysis probability functions soft-question riemann-integration
asked Jan 26 at 15:24
Subhasis BiswasSubhasis Biswas
511411
$endgroup$
$begingroup$
Depends on $F$ and the distribution you put on $F$. In most "natural" ways to do this that I can think of, the probability will be either 0 or 1, depending on whether $F$ consists only of "nice" functions or not.
$endgroup$
– Ian
Jan 26 at 15:25
$begingroup$
does it require the knowledge of measure?
$endgroup$
– Subhasis Biswas
Jan 26 at 15:25
$begingroup$
That also depends on how you do it, but doing anything interesting with this question will probably need some measure theory, yes.
$endgroup$
– Ian
Jan 26 at 15:26
1
$begingroup$
Sort of; the Lebesgue criterion tells you that Riemann integrable functions can't be too far from being continuous in some sense, and there is a vast availability of nowhere continuous functions out there.
$endgroup$
– Ian
Jan 26 at 16:01
1
$begingroup$
If you actually mean that $F$ is the set of all functions from $[a,b]$ to $mathbb{R}$ then there's pretty much no reasonably "uniform-ish" distribution on that which will assign positive probability to the Riemann integrable functions.
$endgroup$
– Ian
Jan 26 at 17:30
|
show 6 more comments
$begingroup$
Suppose we are given a set $F ={f| f:mathbb{R} to mathbb{R}}$, i.e. an arbitrary set (not sure about its countablility) of real-valued functions. What is the probability, that a function $f_k$, chosen at random from the set is Riemann Integrable?
Note: I came up with this problem entirely on my own. So, if there are any "loose-statements", please correct it. I don't have enough knowledge to even approach this. There might be some duplicates out there, which I am not sure of. Is this even a valid question?
Thank you.
real-analysis probability functions soft-question riemann-integration
asked Jan 26 at 15:24
Subhasis BiswasSubhasis Biswas
511411
$endgroup$
Suppose we are given a set $F ={f| f:mathbb{R} to mathbb{R}}$, i.e. an arbitrary set (not sure about its countablility) of real-valued functions. What is the probability, that a function $f_k$, chosen at random from the set is Riemann Integrable?
Note: I came up with this problem entirely on my own. So, if there are any "loose-statements", please correct it. I don't have enough knowledge to even approach this. There might be some duplicates out there, which I am not sure of. Is this even a valid question?
Thank you.
real-analysis probability functions soft-question riemann-integration
real-analysis probability functions soft-question riemann-integration
asked Jan 26 at 15:24
Subhasis BiswasSubhasis Biswas
511411
asked Jan 26 at 15:24
Subhasis BiswasSubhasis Biswas
511411
edited Jan 26 at 15:29
Subhasis Biswas
asked Jan 26 at 15:24
Subhasis BiswasSubhasis Biswas
511411
asked Jan 26 at 15:24
Subhasis BiswasSubhasis Biswas
511411
asked Jan 26 at 15:24
Subhasis BiswasSubhasis Biswas
511411
511411
$begingroup$
Depends on $F$ and the distribution you put on $F$. In most "natural" ways to do this that I can think of, the probability will be either 0 or 1, depending on whether $F$ consists only of "nice" functions or not.
$endgroup$
– Ian
Jan 26 at 15:25
$begingroup$
does it require the knowledge of measure?
$endgroup$
– Subhasis Biswas
Jan 26 at 15:25
$begingroup$
That also depends on how you do it, but doing anything interesting with this question will probably need some measure theory, yes.
$endgroup$
– Ian
Jan 26 at 15:26
1
$begingroup$
Sort of; the Lebesgue criterion tells you that Riemann integrable functions can't be too far from being continuous in some sense, and there is a vast availability of nowhere continuous functions out there.
$endgroup$
– Ian
Jan 26 at 16:01
1
$begingroup$
If you actually mean that $F$ is the set of all functions from $[a,b]$ to $mathbb{R}$ then there's pretty much no reasonably "uniform-ish" distribution on that which will assign positive probability to the Riemann integrable functions.
$endgroup$
– Ian
Jan 26 at 17:30
|
show 6 more comments
$begingroup$
Depends on $F$ and the distribution you put on $F$. In most "natural" ways to do this that I can think of, the probability will be either 0 or 1, depending on whether $F$ consists only of "nice" functions or not.
$endgroup$
– Ian
Jan 26 at 15:25
$begingroup$
does it require the knowledge of measure?
$endgroup$
– Subhasis Biswas
Jan 26 at 15:25
$begingroup$
That also depends on how you do it, but doing anything interesting with this question will probably need some measure theory, yes.
$endgroup$
– Ian
Jan 26 at 15:26
1
$begingroup$
Sort of; the Lebesgue criterion tells you that Riemann integrable functions can't be too far from being continuous in some sense, and there is a vast availability of nowhere continuous functions out there.
$endgroup$
– Ian
Jan 26 at 16:01
1
$begingroup$
If you actually mean that $F$ is the set of all functions from $[a,b]$ to $mathbb{R}$ then there's pretty much no reasonably "uniform-ish" distribution on that which will assign positive probability to the Riemann integrable functions.
$endgroup$
– Ian
Jan 26 at 17:30
$begingroup$
Depends on $F$ and the distribution you put on $F$. In most "natural" ways to do this that I can think of, the probability will be either 0 or 1, depending on whether $F$ consists only of "nice" functions or not.
$endgroup$
– Ian
Jan 26 at 15:25
$begingroup$
Depends on $F$ and the distribution you put on $F$. In most "natural" ways to do this that I can think of, the probability will be either 0 or 1, depending on whether $F$ consists only of "nice" functions or not.
$endgroup$
– Ian
Jan 26 at 15:25
$begingroup$
does it require the knowledge of measure?
$endgroup$
– Subhasis Biswas
Jan 26 at 15:25
$begingroup$
does it require the knowledge of measure?
$endgroup$
– Subhasis Biswas
Jan 26 at 15:25
$begingroup$
That also depends on how you do it, but doing anything interesting with this question will probably need some measure theory, yes.
$endgroup$
– Ian
Jan 26 at 15:26
$begingroup$
That also depends on how you do it, but doing anything interesting with this question will probably need some measure theory, yes.
$endgroup$
– Ian
Jan 26 at 15:26
1
1
$begingroup$
Sort of; the Lebesgue criterion tells you that Riemann integrable functions can't be too far from being continuous in some sense, and there is a vast availability of nowhere continuous functions out there.
$endgroup$
– Ian
Jan 26 at 16:01
$begingroup$
Sort of; the Lebesgue criterion tells you that Riemann integrable functions can't be too far from being continuous in some sense, and there is a vast availability of nowhere continuous functions out there.
$endgroup$
– Ian
Jan 26 at 16:01
1
1
$begingroup$
If you actually mean that $F$ is the set of all functions from $[a,b]$ to $mathbb{R}$ then there's pretty much no reasonably "uniform-ish" distribution on that which will assign positive probability to the Riemann integrable functions.
$endgroup$
– Ian
Jan 26 at 17:30
$begingroup$
If you actually mean that $F$ is the set of all functions from $[a,b]$ to $mathbb{R}$ then there's pretty much no reasonably "uniform-ish" distribution on that which will assign positive probability to the Riemann integrable functions.
$endgroup$
– Ian
Jan 26 at 17:30
|
show 6 more comments
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$begingroup$
Depends on $F$ and the distribution you put on $F$. In most "natural" ways to do this that I can think of, the probability will be either 0 or 1, depending on whether $F$ consists only of "nice" functions or not.
$endgroup$
– Ian
Jan 26 at 15:25
$begingroup$
does it require the knowledge of measure?
$endgroup$
– Subhasis Biswas
Jan 26 at 15:25
$begingroup$
That also depends on how you do it, but doing anything interesting with this question will probably need some measure theory, yes.
$endgroup$
– Ian
Jan 26 at 15:26
1
$begingroup$
Sort of; the Lebesgue criterion tells you that Riemann integrable functions can't be too far from being continuous in some sense, and there is a vast availability of nowhere continuous functions out there.
$endgroup$
– Ian
Jan 26 at 16:01
1
$begingroup$
If you actually mean that $F$ is the set of all functions from $[a,b]$ to $mathbb{R}$ then there's pretty much no reasonably "uniform-ish" distribution on that which will assign positive probability to the Riemann integrable functions.
$endgroup$
– Ian
Jan 26 at 17:30