Mixing
Showing if $f$ unbounded, then $f$ is not Riemann Integrable
$begingroup$
I want to show that if $f:[a,b] to mathbb{R}$ is unbounded, then $f$ is not Riemann Integrable.
I suppose that $f$ is unbounded above:
$$forall_{n in mathbb{N}} exists_{t_n} in [a,b], ; f(t_n) > n$$
and I note that if $f$ is integrable, then there exists two sequences $P, Q < delta$ s.t. $|sigma(f,P,xi) - sigma(f,Q, zeta)| < epsilon$
So if I want to show that if $f$ is not integrable, then I must negate the above statement:
$$exists_{epsilon > 0} forall_{delta > 0} ||P|| < delta ||Q|| < delta, ; |sigma(f,P,xi) - sigma(f,Q,zeta)| geq epsilon$$ where $P$ and $Q$ are sequences in $[a,b]$.
I let $P = Q = left { x_kright }_{k=0}$ and $epsilon = 1$. However, it is here where I am stuck. How do I choose my sequence ${x_k}$?
EDIT: I think I can just pick a sequence with an even interval. i.e, like $x_k = a + kleft(frac{b-a}{m}right )$ where $m$ is the number of subintervals. Then we have that $|Delta x_k| = frac{b-a}{m}$
analysis
asked May 18 '15 at 23:49
QunoQuno
144
$endgroup$
add a comment |
$begingroup$
I want to show that if $f:[a,b] to mathbb{R}$ is unbounded, then $f$ is not Riemann Integrable.
I suppose that $f$ is unbounded above:
$$forall_{n in mathbb{N}} exists_{t_n} in [a,b], ; f(t_n) > n$$
and I note that if $f$ is integrable, then there exists two sequences $P, Q < delta$ s.t. $|sigma(f,P,xi) - sigma(f,Q, zeta)| < epsilon$
So if I want to show that if $f$ is not integrable, then I must negate the above statement:
$$exists_{epsilon > 0} forall_{delta > 0} ||P|| < delta ||Q|| < delta, ; |sigma(f,P,xi) - sigma(f,Q,zeta)| geq epsilon$$ where $P$ and $Q$ are sequences in $[a,b]$.
I let $P = Q = left { x_kright }_{k=0}$ and $epsilon = 1$. However, it is here where I am stuck. How do I choose my sequence ${x_k}$?
EDIT: I think I can just pick a sequence with an even interval. i.e, like $x_k = a + kleft(frac{b-a}{m}right )$ where $m$ is the number of subintervals. Then we have that $|Delta x_k| = frac{b-a}{m}$
analysis
asked May 18 '15 at 23:49
QunoQuno
144
$endgroup$
$begingroup$
You can create a sequence of tagged partitions where the Riemann sum becomes arbitrarily large, i.e. $(x_{i+1}-x_i)f(x_i^*)$can be made arbitrarily large if $f$ is unbounded on $[x_i,x_{i+1}]$.
$endgroup$
– yoyo
May 19 '15 at 0:09
add a comment |
$begingroup$
I want to show that if $f:[a,b] to mathbb{R}$ is unbounded, then $f$ is not Riemann Integrable.
I suppose that $f$ is unbounded above:
$$forall_{n in mathbb{N}} exists_{t_n} in [a,b], ; f(t_n) > n$$
and I note that if $f$ is integrable, then there exists two sequences $P, Q < delta$ s.t. $|sigma(f,P,xi) - sigma(f,Q, zeta)| < epsilon$
So if I want to show that if $f$ is not integrable, then I must negate the above statement:
$$exists_{epsilon > 0} forall_{delta > 0} ||P|| < delta ||Q|| < delta, ; |sigma(f,P,xi) - sigma(f,Q,zeta)| geq epsilon$$ where $P$ and $Q$ are sequences in $[a,b]$.
I let $P = Q = left { x_kright }_{k=0}$ and $epsilon = 1$. However, it is here where I am stuck. How do I choose my sequence ${x_k}$?
EDIT: I think I can just pick a sequence with an even interval. i.e, like $x_k = a + kleft(frac{b-a}{m}right )$ where $m$ is the number of subintervals. Then we have that $|Delta x_k| = frac{b-a}{m}$
analysis
asked May 18 '15 at 23:49
QunoQuno
144
$endgroup$
I want to show that if $f:[a,b] to mathbb{R}$ is unbounded, then $f$ is not Riemann Integrable.
I suppose that $f$ is unbounded above:
$$forall_{n in mathbb{N}} exists_{t_n} in [a,b], ; f(t_n) > n$$
and I note that if $f$ is integrable, then there exists two sequences $P, Q < delta$ s.t. $|sigma(f,P,xi) - sigma(f,Q, zeta)| < epsilon$
So if I want to show that if $f$ is not integrable, then I must negate the above statement:
$$exists_{epsilon > 0} forall_{delta > 0} ||P|| < delta ||Q|| < delta, ; |sigma(f,P,xi) - sigma(f,Q,zeta)| geq epsilon$$ where $P$ and $Q$ are sequences in $[a,b]$.
I let $P = Q = left { x_kright }_{k=0}$ and $epsilon = 1$. However, it is here where I am stuck. How do I choose my sequence ${x_k}$?
EDIT: I think I can just pick a sequence with an even interval. i.e, like $x_k = a + kleft(frac{b-a}{m}right )$ where $m$ is the number of subintervals. Then we have that $|Delta x_k| = frac{b-a}{m}$
analysis
analysis
asked May 18 '15 at 23:49
QunoQuno
144
asked May 18 '15 at 23:49
QunoQuno
144
edited May 19 '15 at 0:14
Quno
asked May 18 '15 at 23:49
QunoQuno
144
asked May 18 '15 at 23:49
QunoQuno
144
asked May 18 '15 at 23:49
QunoQuno
144
144
$begingroup$
You can create a sequence of tagged partitions where the Riemann sum becomes arbitrarily large, i.e. $(x_{i+1}-x_i)f(x_i^*)$can be made arbitrarily large if $f$ is unbounded on $[x_i,x_{i+1}]$.
$endgroup$
– yoyo
May 19 '15 at 0:09
add a comment |
$begingroup$
You can create a sequence of tagged partitions where the Riemann sum becomes arbitrarily large, i.e. $(x_{i+1}-x_i)f(x_i^*)$can be made arbitrarily large if $f$ is unbounded on $[x_i,x_{i+1}]$.
$endgroup$
– yoyo
May 19 '15 at 0:09
$begingroup$
You can create a sequence of tagged partitions where the Riemann sum becomes arbitrarily large, i.e. $(x_{i+1}-x_i)f(x_i^*)$can be made arbitrarily large if $f$ is unbounded on $[x_i,x_{i+1}]$.
$endgroup$
– yoyo
May 19 '15 at 0:09
$begingroup$
You can create a sequence of tagged partitions where the Riemann sum becomes arbitrarily large, i.e. $(x_{i+1}-x_i)f(x_i^*)$can be made arbitrarily large if $f$ is unbounded on $[x_i,x_{i+1}]$.
$endgroup$
– yoyo
May 19 '15 at 0:09
add a comment |
1 Answer
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$begingroup$
We can assume that $f$ is unbounded above without loss of generality.
Then there are $x_n$ such that $f(x_n) ge n$, and since $[a,b]$ is compact,
there is some accumulation point $c$ such that $sup_{x in (c-epsilon, c+epsilon)cap[a,b]} f(x) = infty$ for all $epsilon>0$.
Let $P$ be a partition of $[a,b]$. Suppose $c in I^circ$, where $I$ is an interval of the partition. Then we see that
$sup_{x in I} f(x) = infty$.
We see that if $c$ is an endpoint, we obtain the same result.
If $c$ lies on the boundeary between two intervals, at least one of the intervals must satisfy $sup_{x in I} f(x) = infty$.
In all cases, we see that $U(f,P) = infty$, hence $f$ is not integrable.
answered May 19 '15 at 0:29
copper.hatcopper.hat
127k559160
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add a comment |
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$begingroup$
We can assume that $f$ is unbounded above without loss of generality.
Then there are $x_n$ such that $f(x_n) ge n$, and since $[a,b]$ is compact,
there is some accumulation point $c$ such that $sup_{x in (c-epsilon, c+epsilon)cap[a,b]} f(x) = infty$ for all $epsilon>0$.
Let $P$ be a partition of $[a,b]$. Suppose $c in I^circ$, where $I$ is an interval of the partition. Then we see that
$sup_{x in I} f(x) = infty$.
We see that if $c$ is an endpoint, we obtain the same result.
If $c$ lies on the boundeary between two intervals, at least one of the intervals must satisfy $sup_{x in I} f(x) = infty$.
In all cases, we see that $U(f,P) = infty$, hence $f$ is not integrable.
answered May 19 '15 at 0:29
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
We can assume that $f$ is unbounded above without loss of generality.
Then there are $x_n$ such that $f(x_n) ge n$, and since $[a,b]$ is compact,
there is some accumulation point $c$ such that $sup_{x in (c-epsilon, c+epsilon)cap[a,b]} f(x) = infty$ for all $epsilon>0$.
Let $P$ be a partition of $[a,b]$. Suppose $c in I^circ$, where $I$ is an interval of the partition. Then we see that
$sup_{x in I} f(x) = infty$.
We see that if $c$ is an endpoint, we obtain the same result.
If $c$ lies on the boundeary between two intervals, at least one of the intervals must satisfy $sup_{x in I} f(x) = infty$.
In all cases, we see that $U(f,P) = infty$, hence $f$ is not integrable.
answered May 19 '15 at 0:29
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
We can assume that $f$ is unbounded above without loss of generality.
Then there are $x_n$ such that $f(x_n) ge n$, and since $[a,b]$ is compact,
there is some accumulation point $c$ such that $sup_{x in (c-epsilon, c+epsilon)cap[a,b]} f(x) = infty$ for all $epsilon>0$.
Let $P$ be a partition of $[a,b]$. Suppose $c in I^circ$, where $I$ is an interval of the partition. Then we see that
$sup_{x in I} f(x) = infty$.
We see that if $c$ is an endpoint, we obtain the same result.
If $c$ lies on the boundeary between two intervals, at least one of the intervals must satisfy $sup_{x in I} f(x) = infty$.
In all cases, we see that $U(f,P) = infty$, hence $f$ is not integrable.
answered May 19 '15 at 0:29
copper.hatcopper.hat
127k559160
$endgroup$
We can assume that $f$ is unbounded above without loss of generality.
Then there are $x_n$ such that $f(x_n) ge n$, and since $[a,b]$ is compact,
there is some accumulation point $c$ such that $sup_{x in (c-epsilon, c+epsilon)cap[a,b]} f(x) = infty$ for all $epsilon>0$.
Let $P$ be a partition of $[a,b]$. Suppose $c in I^circ$, where $I$ is an interval of the partition. Then we see that
$sup_{x in I} f(x) = infty$.
We see that if $c$ is an endpoint, we obtain the same result.
If $c$ lies on the boundeary between two intervals, at least one of the intervals must satisfy $sup_{x in I} f(x) = infty$.
In all cases, we see that $U(f,P) = infty$, hence $f$ is not integrable.
answered May 19 '15 at 0:29
copper.hatcopper.hat
127k559160
answered May 19 '15 at 0:29
copper.hatcopper.hat
127k559160
answered May 19 '15 at 0:29
copper.hatcopper.hat
127k559160
answered May 19 '15 at 0:29
copper.hatcopper.hat
127k559160
127k559160
add a comment |
add a comment |
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$begingroup$
You can create a sequence of tagged partitions where the Riemann sum becomes arbitrarily large, i.e. $(x_{i+1}-x_i)f(x_i^*)$can be made arbitrarily large if $f$ is unbounded on $[x_i,x_{i+1}]$.
$endgroup$
– yoyo
May 19 '15 at 0:09