Mixing
Riemann Integral: prove ${U(P,f):P$ is a partition of $[a,b]}$ is bounded below
$begingroup$
[Defining the Riemann Integral: ]
We consider a partition, $P={a=x_0<x_1<...<x_n=b}$ of $[a,b]$ and a bounded function $f:[a,b]rightarrowmathbb{R}$.
Next, we define-
$$M_i=sup{f(x)|xin[x_{i-1},x_i]}$$
and
$$U(P,f)=underset{i=1}{overset{n}{sum}}M_i(x_i-x_{i-1})$$
What was also mentioned was $A_1={U(P,f):P$ is a partition of $[a,b]}$ is bounded below.
Intuitively, of course, the actual area covered by $f(x)le A_1$, which is why we can say that $A_1$ is bounded below. But, how do I formally prove that?
riemann-integration upper-lower-bounds
asked Jan 6 at 14:31
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
[Defining the Riemann Integral: ]
We consider a partition, $P={a=x_0<x_1<...<x_n=b}$ of $[a,b]$ and a bounded function $f:[a,b]rightarrowmathbb{R}$.
Next, we define-
$$M_i=sup{f(x)|xin[x_{i-1},x_i]}$$
and
$$U(P,f)=underset{i=1}{overset{n}{sum}}M_i(x_i-x_{i-1})$$
What was also mentioned was $A_1={U(P,f):P$ is a partition of $[a,b]}$ is bounded below.
Intuitively, of course, the actual area covered by $f(x)le A_1$, which is why we can say that $A_1$ is bounded below. But, how do I formally prove that?
riemann-integration upper-lower-bounds
asked Jan 6 at 14:31
Za IraZa Ira
161115
$endgroup$
1
$begingroup$
$f$ is bounded, so the upper sum over an interval is bounded below by the lower bound on $f$ times the length of the interval.
$endgroup$
– jgon
Jan 6 at 14:35
$begingroup$
@jgon do you mean $U(P,f)$ is bounded below by $[min(M_i)]*(b-a)$?
$endgroup$
– Za Ira
Jan 6 at 14:41
1
$begingroup$
Well no, the min of the $M_i$ depends on the partition. I mean that you know that $|f|<M$, so you know $f>-M$ for some $M$
$endgroup$
– jgon
Jan 6 at 14:47
$begingroup$
@jgon what is that $M$?
$endgroup$
– Za Ira
Jan 6 at 14:49
add a comment |
$begingroup$
[Defining the Riemann Integral: ]
We consider a partition, $P={a=x_0<x_1<...<x_n=b}$ of $[a,b]$ and a bounded function $f:[a,b]rightarrowmathbb{R}$.
Next, we define-
$$M_i=sup{f(x)|xin[x_{i-1},x_i]}$$
and
$$U(P,f)=underset{i=1}{overset{n}{sum}}M_i(x_i-x_{i-1})$$
What was also mentioned was $A_1={U(P,f):P$ is a partition of $[a,b]}$ is bounded below.
Intuitively, of course, the actual area covered by $f(x)le A_1$, which is why we can say that $A_1$ is bounded below. But, how do I formally prove that?
riemann-integration upper-lower-bounds
asked Jan 6 at 14:31
Za IraZa Ira
161115
$endgroup$
[Defining the Riemann Integral: ]
We consider a partition, $P={a=x_0<x_1<...<x_n=b}$ of $[a,b]$ and a bounded function $f:[a,b]rightarrowmathbb{R}$.
Next, we define-
$$M_i=sup{f(x)|xin[x_{i-1},x_i]}$$
and
$$U(P,f)=underset{i=1}{overset{n}{sum}}M_i(x_i-x_{i-1})$$
What was also mentioned was $A_1={U(P,f):P$ is a partition of $[a,b]}$ is bounded below.
Intuitively, of course, the actual area covered by $f(x)le A_1$, which is why we can say that $A_1$ is bounded below. But, how do I formally prove that?
riemann-integration upper-lower-bounds
riemann-integration upper-lower-bounds
asked Jan 6 at 14:31
Za IraZa Ira
161115
asked Jan 6 at 14:31
Za IraZa Ira
161115
asked Jan 6 at 14:31
Za IraZa Ira
161115
asked Jan 6 at 14:31
Za IraZa Ira
161115
asked Jan 6 at 14:31
Za IraZa Ira
161115
161115
1
$begingroup$
$f$ is bounded, so the upper sum over an interval is bounded below by the lower bound on $f$ times the length of the interval.
$endgroup$
– jgon
Jan 6 at 14:35
$begingroup$
@jgon do you mean $U(P,f)$ is bounded below by $[min(M_i)]*(b-a)$?
$endgroup$
– Za Ira
Jan 6 at 14:41
1
$begingroup$
Well no, the min of the $M_i$ depends on the partition. I mean that you know that $|f|<M$, so you know $f>-M$ for some $M$
$endgroup$
– jgon
Jan 6 at 14:47
$begingroup$
@jgon what is that $M$?
$endgroup$
– Za Ira
Jan 6 at 14:49
add a comment |
1
$begingroup$
$f$ is bounded, so the upper sum over an interval is bounded below by the lower bound on $f$ times the length of the interval.
$endgroup$
– jgon
Jan 6 at 14:35
$begingroup$
@jgon do you mean $U(P,f)$ is bounded below by $[min(M_i)]*(b-a)$?
$endgroup$
– Za Ira
Jan 6 at 14:41
1
$begingroup$
Well no, the min of the $M_i$ depends on the partition. I mean that you know that $|f|<M$, so you know $f>-M$ for some $M$
$endgroup$
– jgon
Jan 6 at 14:47
$begingroup$
@jgon what is that $M$?
$endgroup$
– Za Ira
Jan 6 at 14:49
1
1
$begingroup$
$f$ is bounded, so the upper sum over an interval is bounded below by the lower bound on $f$ times the length of the interval.
$endgroup$
– jgon
Jan 6 at 14:35
$begingroup$
$f$ is bounded, so the upper sum over an interval is bounded below by the lower bound on $f$ times the length of the interval.
$endgroup$
– jgon
Jan 6 at 14:35
$begingroup$
@jgon do you mean $U(P,f)$ is bounded below by $[min(M_i)]*(b-a)$?
$endgroup$
– Za Ira
Jan 6 at 14:41
$begingroup$
@jgon do you mean $U(P,f)$ is bounded below by $[min(M_i)]*(b-a)$?
$endgroup$
– Za Ira
Jan 6 at 14:41
1
1
$begingroup$
Well no, the min of the $M_i$ depends on the partition. I mean that you know that $|f|<M$, so you know $f>-M$ for some $M$
$endgroup$
– jgon
Jan 6 at 14:47
$begingroup$
Well no, the min of the $M_i$ depends on the partition. I mean that you know that $|f|<M$, so you know $f>-M$ for some $M$
$endgroup$
– jgon
Jan 6 at 14:47
$begingroup$
@jgon what is that $M$?
$endgroup$
– Za Ira
Jan 6 at 14:49
$begingroup$
@jgon what is that $M$?
$endgroup$
– Za Ira
Jan 6 at 14:49
add a comment |
2 Answers
2
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oldest
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$begingroup$
Suppose $P$ is a partition and that $Q$ is the partition obtained by $P$ by adding a single point. Prove that $L(P,f)le L(Q,f)$, where $L$ denotes the lower sum.
Since any partition $P$ can be obtained from the trivial partition $P_0={a,b}$ by successively adding points, induction shows that $L(P_0,f)le L(P,f)$.
Next observe that $L(P,f)le U(P,f)$.
Similarly, lower sums are upper bounded.
It is perhaps simpler to observe that
$$
sum_{i=1}^n M_i(x_i-x_{i-1})gesum_{i=1}^n m(x_i-x_{i-1})=m(b-a)
$$
where $m=inf{f(x):xin[a,b]}$, but the above approach allows for going on with the construction of the Riemann integral.
answered Jan 6 at 14:51
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
egreg's answer is excellent (+1), but I thought I'd elaborate on my comment.
We are given that $f$ is bounded, which by definition means that there exists $MinBbb{R}$, $M>0$ such that
$$|f(x)|<M$$
for $xin[a,b]$.
Then $f(x)>-M$ for all $xin[a,b]$, and therefore
$U(P,f) > -M(b-a)$.
answered Jan 6 at 15:12
jgonjgon
13.6k22041
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Suppose $P$ is a partition and that $Q$ is the partition obtained by $P$ by adding a single point. Prove that $L(P,f)le L(Q,f)$, where $L$ denotes the lower sum.
Since any partition $P$ can be obtained from the trivial partition $P_0={a,b}$ by successively adding points, induction shows that $L(P_0,f)le L(P,f)$.
Next observe that $L(P,f)le U(P,f)$.
Similarly, lower sums are upper bounded.
It is perhaps simpler to observe that
$$
sum_{i=1}^n M_i(x_i-x_{i-1})gesum_{i=1}^n m(x_i-x_{i-1})=m(b-a)
$$
where $m=inf{f(x):xin[a,b]}$, but the above approach allows for going on with the construction of the Riemann integral.
answered Jan 6 at 14:51
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Suppose $P$ is a partition and that $Q$ is the partition obtained by $P$ by adding a single point. Prove that $L(P,f)le L(Q,f)$, where $L$ denotes the lower sum.
Since any partition $P$ can be obtained from the trivial partition $P_0={a,b}$ by successively adding points, induction shows that $L(P_0,f)le L(P,f)$.
Next observe that $L(P,f)le U(P,f)$.
Similarly, lower sums are upper bounded.
It is perhaps simpler to observe that
$$
sum_{i=1}^n M_i(x_i-x_{i-1})gesum_{i=1}^n m(x_i-x_{i-1})=m(b-a)
$$
where $m=inf{f(x):xin[a,b]}$, but the above approach allows for going on with the construction of the Riemann integral.
answered Jan 6 at 14:51
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Suppose $P$ is a partition and that $Q$ is the partition obtained by $P$ by adding a single point. Prove that $L(P,f)le L(Q,f)$, where $L$ denotes the lower sum.
Since any partition $P$ can be obtained from the trivial partition $P_0={a,b}$ by successively adding points, induction shows that $L(P_0,f)le L(P,f)$.
Next observe that $L(P,f)le U(P,f)$.
Similarly, lower sums are upper bounded.
It is perhaps simpler to observe that
$$
sum_{i=1}^n M_i(x_i-x_{i-1})gesum_{i=1}^n m(x_i-x_{i-1})=m(b-a)
$$
where $m=inf{f(x):xin[a,b]}$, but the above approach allows for going on with the construction of the Riemann integral.
answered Jan 6 at 14:51
egregegreg
180k1485202
$endgroup$
Suppose $P$ is a partition and that $Q$ is the partition obtained by $P$ by adding a single point. Prove that $L(P,f)le L(Q,f)$, where $L$ denotes the lower sum.
Since any partition $P$ can be obtained from the trivial partition $P_0={a,b}$ by successively adding points, induction shows that $L(P_0,f)le L(P,f)$.
Next observe that $L(P,f)le U(P,f)$.
Similarly, lower sums are upper bounded.
It is perhaps simpler to observe that
$$
sum_{i=1}^n M_i(x_i-x_{i-1})gesum_{i=1}^n m(x_i-x_{i-1})=m(b-a)
$$
where $m=inf{f(x):xin[a,b]}$, but the above approach allows for going on with the construction of the Riemann integral.
answered Jan 6 at 14:51
egregegreg
180k1485202
edited Jan 6 at 14:57
answered Jan 6 at 14:51
egregegreg
180k1485202
answered Jan 6 at 14:51
egregegreg
180k1485202
answered Jan 6 at 14:51
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
$begingroup$
egreg's answer is excellent (+1), but I thought I'd elaborate on my comment.
We are given that $f$ is bounded, which by definition means that there exists $MinBbb{R}$, $M>0$ such that
$$|f(x)|<M$$
for $xin[a,b]$.
Then $f(x)>-M$ for all $xin[a,b]$, and therefore
$U(P,f) > -M(b-a)$.
answered Jan 6 at 15:12
jgonjgon
13.6k22041
$endgroup$
add a comment |
$begingroup$
egreg's answer is excellent (+1), but I thought I'd elaborate on my comment.
We are given that $f$ is bounded, which by definition means that there exists $MinBbb{R}$, $M>0$ such that
$$|f(x)|<M$$
for $xin[a,b]$.
Then $f(x)>-M$ for all $xin[a,b]$, and therefore
$U(P,f) > -M(b-a)$.
answered Jan 6 at 15:12
jgonjgon
13.6k22041
$endgroup$
add a comment |
$begingroup$
egreg's answer is excellent (+1), but I thought I'd elaborate on my comment.
We are given that $f$ is bounded, which by definition means that there exists $MinBbb{R}$, $M>0$ such that
$$|f(x)|<M$$
for $xin[a,b]$.
Then $f(x)>-M$ for all $xin[a,b]$, and therefore
$U(P,f) > -M(b-a)$.
answered Jan 6 at 15:12
jgonjgon
13.6k22041
$endgroup$
egreg's answer is excellent (+1), but I thought I'd elaborate on my comment.
We are given that $f$ is bounded, which by definition means that there exists $MinBbb{R}$, $M>0$ such that
$$|f(x)|<M$$
for $xin[a,b]$.
Then $f(x)>-M$ for all $xin[a,b]$, and therefore
$U(P,f) > -M(b-a)$.
answered Jan 6 at 15:12
jgonjgon
13.6k22041
answered Jan 6 at 15:12
jgonjgon
13.6k22041
answered Jan 6 at 15:12
jgonjgon
13.6k22041
answered Jan 6 at 15:12
jgonjgon
13.6k22041
13.6k22041
add a comment |
add a comment |
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$begingroup$
$f$ is bounded, so the upper sum over an interval is bounded below by the lower bound on $f$ times the length of the interval.
$endgroup$
– jgon
Jan 6 at 14:35
$begingroup$
@jgon do you mean $U(P,f)$ is bounded below by $[min(M_i)]*(b-a)$?
$endgroup$
– Za Ira
Jan 6 at 14:41
1
$begingroup$
Well no, the min of the $M_i$ depends on the partition. I mean that you know that $|f|<M$, so you know $f>-M$ for some $M$
$endgroup$
– jgon
Jan 6 at 14:47
$begingroup$
@jgon what is that $M$?
$endgroup$
– Za Ira
Jan 6 at 14:49