Upper and lower Riemann sum of Partitions depending on $n$












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I am currently self studying Riemann integrability and got to a part that states that if partitions P and Q are such that Q is a refinement of P then $$L (f,P) leq U (f,Q) leq U (f,Q) leq U (f,P)$$
I fully understand the proof. It is based on the fact that every point of P is in Q.



My thought is this: Suppose I have a partition dependent on n, say:
$$P_n:={x_0,x_{1,n},x_{2,n},...,x_{p,n}}$$
Like for example,
$$P_n:={0,frac{1}{n},frac{2}{n},...,1}$$
Will I also have that
$$L (f,P_n) leq U (f,P_{n+1}) leq U (f,P_{n+1}) leq U (f,P_n)?$$
I am stuck on how to approach this, or even get a counter example.



If it is false, should there be any condition to be put on $f$, say continuity or something else?










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  • $begingroup$
    I believe you mea $L leq L leq U leq U.$ If you have $P_n$ is a subfamily of $P_{n+1}$ then it's a corollary of what you have already.
    $endgroup$
    – Will M.
    Jan 25 at 20:56










  • $begingroup$
    @WillM. The problem is that $P_{n+1}$ is not necessarily a refinement of $P_n$
    $endgroup$
    – user397197
    Jan 25 at 21:11












  • $begingroup$
    I am silly, I forgot the following result before: if $P$ and $Q$ are any partitions, then $L(f, P) leq U(f, Q).$ (Use what you have with a common refinement.)
    $endgroup$
    – Will M.
    Jan 25 at 22:32
















0












$begingroup$


I am currently self studying Riemann integrability and got to a part that states that if partitions P and Q are such that Q is a refinement of P then $$L (f,P) leq U (f,Q) leq U (f,Q) leq U (f,P)$$
I fully understand the proof. It is based on the fact that every point of P is in Q.



My thought is this: Suppose I have a partition dependent on n, say:
$$P_n:={x_0,x_{1,n},x_{2,n},...,x_{p,n}}$$
Like for example,
$$P_n:={0,frac{1}{n},frac{2}{n},...,1}$$
Will I also have that
$$L (f,P_n) leq U (f,P_{n+1}) leq U (f,P_{n+1}) leq U (f,P_n)?$$
I am stuck on how to approach this, or even get a counter example.



If it is false, should there be any condition to be put on $f$, say continuity or something else?










share|cite|improve this question











$endgroup$












  • $begingroup$
    I believe you mea $L leq L leq U leq U.$ If you have $P_n$ is a subfamily of $P_{n+1}$ then it's a corollary of what you have already.
    $endgroup$
    – Will M.
    Jan 25 at 20:56










  • $begingroup$
    @WillM. The problem is that $P_{n+1}$ is not necessarily a refinement of $P_n$
    $endgroup$
    – user397197
    Jan 25 at 21:11












  • $begingroup$
    I am silly, I forgot the following result before: if $P$ and $Q$ are any partitions, then $L(f, P) leq U(f, Q).$ (Use what you have with a common refinement.)
    $endgroup$
    – Will M.
    Jan 25 at 22:32














0












0








0





$begingroup$


I am currently self studying Riemann integrability and got to a part that states that if partitions P and Q are such that Q is a refinement of P then $$L (f,P) leq U (f,Q) leq U (f,Q) leq U (f,P)$$
I fully understand the proof. It is based on the fact that every point of P is in Q.



My thought is this: Suppose I have a partition dependent on n, say:
$$P_n:={x_0,x_{1,n},x_{2,n},...,x_{p,n}}$$
Like for example,
$$P_n:={0,frac{1}{n},frac{2}{n},...,1}$$
Will I also have that
$$L (f,P_n) leq U (f,P_{n+1}) leq U (f,P_{n+1}) leq U (f,P_n)?$$
I am stuck on how to approach this, or even get a counter example.



If it is false, should there be any condition to be put on $f$, say continuity or something else?










share|cite|improve this question











$endgroup$




I am currently self studying Riemann integrability and got to a part that states that if partitions P and Q are such that Q is a refinement of P then $$L (f,P) leq U (f,Q) leq U (f,Q) leq U (f,P)$$
I fully understand the proof. It is based on the fact that every point of P is in Q.



My thought is this: Suppose I have a partition dependent on n, say:
$$P_n:={x_0,x_{1,n},x_{2,n},...,x_{p,n}}$$
Like for example,
$$P_n:={0,frac{1}{n},frac{2}{n},...,1}$$
Will I also have that
$$L (f,P_n) leq U (f,P_{n+1}) leq U (f,P_{n+1}) leq U (f,P_n)?$$
I am stuck on how to approach this, or even get a counter example.



If it is false, should there be any condition to be put on $f$, say continuity or something else?







real-analysis analysis riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 20:51

























asked Jan 25 at 20:44







user397197



















  • $begingroup$
    I believe you mea $L leq L leq U leq U.$ If you have $P_n$ is a subfamily of $P_{n+1}$ then it's a corollary of what you have already.
    $endgroup$
    – Will M.
    Jan 25 at 20:56










  • $begingroup$
    @WillM. The problem is that $P_{n+1}$ is not necessarily a refinement of $P_n$
    $endgroup$
    – user397197
    Jan 25 at 21:11












  • $begingroup$
    I am silly, I forgot the following result before: if $P$ and $Q$ are any partitions, then $L(f, P) leq U(f, Q).$ (Use what you have with a common refinement.)
    $endgroup$
    – Will M.
    Jan 25 at 22:32


















  • $begingroup$
    I believe you mea $L leq L leq U leq U.$ If you have $P_n$ is a subfamily of $P_{n+1}$ then it's a corollary of what you have already.
    $endgroup$
    – Will M.
    Jan 25 at 20:56










  • $begingroup$
    @WillM. The problem is that $P_{n+1}$ is not necessarily a refinement of $P_n$
    $endgroup$
    – user397197
    Jan 25 at 21:11












  • $begingroup$
    I am silly, I forgot the following result before: if $P$ and $Q$ are any partitions, then $L(f, P) leq U(f, Q).$ (Use what you have with a common refinement.)
    $endgroup$
    – Will M.
    Jan 25 at 22:32
















$begingroup$
I believe you mea $L leq L leq U leq U.$ If you have $P_n$ is a subfamily of $P_{n+1}$ then it's a corollary of what you have already.
$endgroup$
– Will M.
Jan 25 at 20:56




$begingroup$
I believe you mea $L leq L leq U leq U.$ If you have $P_n$ is a subfamily of $P_{n+1}$ then it's a corollary of what you have already.
$endgroup$
– Will M.
Jan 25 at 20:56












$begingroup$
@WillM. The problem is that $P_{n+1}$ is not necessarily a refinement of $P_n$
$endgroup$
– user397197
Jan 25 at 21:11






$begingroup$
@WillM. The problem is that $P_{n+1}$ is not necessarily a refinement of $P_n$
$endgroup$
– user397197
Jan 25 at 21:11














$begingroup$
I am silly, I forgot the following result before: if $P$ and $Q$ are any partitions, then $L(f, P) leq U(f, Q).$ (Use what you have with a common refinement.)
$endgroup$
– Will M.
Jan 25 at 22:32




$begingroup$
I am silly, I forgot the following result before: if $P$ and $Q$ are any partitions, then $L(f, P) leq U(f, Q).$ (Use what you have with a common refinement.)
$endgroup$
– Will M.
Jan 25 at 22:32










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