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Proof verification of monotone functions $f: [a, b] rightarrow mathbbR$ are integrable
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I am reading a calculus textbook and I am having trouble reading a proof that shows that all monotone functions $f: [a, b] rightarrow mathbbR$ are integrable. This proof is not complete, specifically the textbook wants me to complete the proof as an exercise:
Without loss of generality let $f$ be monotone increasing. We'll
divide the interval $[f(a), f(b)]$ up into $n$ equidistant
subintervals with length $$h = fracf(b) - f(a)n$$ and $y_k :=
f(a) + kh$ for $k = 0, 1, dots, n$. Furthermore, let $a_0 := a$ and
$$a_k := supx in [a,b] : f(x) leq y_k$$ for $k = 1, dots, n$.
Because $f$ is monotone, it follows that $$a = a_0 leq a_1 leq dots
leq a_n = b$$ Furthermore, we have (EXERCISE) $$x in (a_k-1, a_k) Rightarrow f(x) in ( y_k-1, y_k]$$
$dots$
I am really unsure if what I did is correct:
If $x < a_k$, then $f(x) leq f(a_k) leq y_k$, because $f$ is monotone increasing. So $f(x) leq y_k$.
If $x > a_k-1$, then there can be no such $x$ with $f(x) leq y_k-1$ because that would contradict the supremum property of $a_k-1$. So we have $y_k-1 < f(x)$.
So $$x in (a_k-1, a_k) Rightarrow f(x) in ( y_k-1, y_k]$$
calculus proof-verification
asked Aug 6 at 9:07
PhysicsUndergraduateStudent
527
add a comment |Â
up vote
2
down vote
favorite
I am reading a calculus textbook and I am having trouble reading a proof that shows that all monotone functions $f: [a, b] rightarrow mathbbR$ are integrable. This proof is not complete, specifically the textbook wants me to complete the proof as an exercise:
Without loss of generality let $f$ be monotone increasing. We'll
divide the interval $[f(a), f(b)]$ up into $n$ equidistant
subintervals with length $$h = fracf(b) - f(a)n$$ and $y_k :=
f(a) + kh$ for $k = 0, 1, dots, n$. Furthermore, let $a_0 := a$ and
$$a_k := supx in [a,b] : f(x) leq y_k$$ for $k = 1, dots, n$.
Because $f$ is monotone, it follows that $$a = a_0 leq a_1 leq dots
leq a_n = b$$ Furthermore, we have (EXERCISE) $$x in (a_k-1, a_k) Rightarrow f(x) in ( y_k-1, y_k]$$
$dots$
I am really unsure if what I did is correct:
If $x < a_k$, then $f(x) leq f(a_k) leq y_k$, because $f$ is monotone increasing. So $f(x) leq y_k$.
If $x > a_k-1$, then there can be no such $x$ with $f(x) leq y_k-1$ because that would contradict the supremum property of $a_k-1$. So we have $y_k-1 < f(x)$.
So $$x in (a_k-1, a_k) Rightarrow f(x) in ( y_k-1, y_k]$$
calculus proof-verification
asked Aug 6 at 9:07
PhysicsUndergraduateStudent
527
i don't see how $f(x) le y_k-1$ contradicts the supremum property of $a_k-1$. in fact the proof is much easier than you might think.
– pointguard0
Aug 6 at 9:28
@pointguard0 Because of the definition of $a_k-1$, it is the "largest" $x$ for which we have $f(x) leq y_k-1$. Any $x$ greater than $a_k-1$ does no longer suffice the inequality $f(x) leq y_k-1$. Or am I missing something?
– PhysicsUndergraduateStudent
Aug 8 at 13:46
and hence, $f(x) > y_k-1$ for all $x > a_k-1$. this is as simple as that
– pointguard0
Aug 8 at 14:06
@pointguard0 Yes, that's exactly what I wrote in my original post.
– PhysicsUndergraduateStudent
Aug 8 at 14:08
okay then, sorry, if i misunderstood. upvoted :)
– pointguard0
Aug 8 at 14:09
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am reading a calculus textbook and I am having trouble reading a proof that shows that all monotone functions $f: [a, b] rightarrow mathbbR$ are integrable. This proof is not complete, specifically the textbook wants me to complete the proof as an exercise:
Without loss of generality let $f$ be monotone increasing. We'll
divide the interval $[f(a), f(b)]$ up into $n$ equidistant
subintervals with length $$h = fracf(b) - f(a)n$$ and $y_k :=
f(a) + kh$ for $k = 0, 1, dots, n$. Furthermore, let $a_0 := a$ and
$$a_k := supx in [a,b] : f(x) leq y_k$$ for $k = 1, dots, n$.
Because $f$ is monotone, it follows that $$a = a_0 leq a_1 leq dots
leq a_n = b$$ Furthermore, we have (EXERCISE) $$x in (a_k-1, a_k) Rightarrow f(x) in ( y_k-1, y_k]$$
$dots$
I am really unsure if what I did is correct:
If $x < a_k$, then $f(x) leq f(a_k) leq y_k$, because $f$ is monotone increasing. So $f(x) leq y_k$.
If $x > a_k-1$, then there can be no such $x$ with $f(x) leq y_k-1$ because that would contradict the supremum property of $a_k-1$. So we have $y_k-1 < f(x)$.
So $$x in (a_k-1, a_k) Rightarrow f(x) in ( y_k-1, y_k]$$
calculus proof-verification
asked Aug 6 at 9:07
PhysicsUndergraduateStudent
527
I am reading a calculus textbook and I am having trouble reading a proof that shows that all monotone functions $f: [a, b] rightarrow mathbbR$ are integrable. This proof is not complete, specifically the textbook wants me to complete the proof as an exercise:
Without loss of generality let $f$ be monotone increasing. We'll
divide the interval $[f(a), f(b)]$ up into $n$ equidistant
subintervals with length $$h = fracf(b) - f(a)n$$ and $y_k :=
f(a) + kh$ for $k = 0, 1, dots, n$. Furthermore, let $a_0 := a$ and
$$a_k := supx in [a,b] : f(x) leq y_k$$ for $k = 1, dots, n$.
Because $f$ is monotone, it follows that $$a = a_0 leq a_1 leq dots
leq a_n = b$$ Furthermore, we have (EXERCISE) $$x in (a_k-1, a_k) Rightarrow f(x) in ( y_k-1, y_k]$$
$dots$
I am really unsure if what I did is correct:
If $x < a_k$, then $f(x) leq f(a_k) leq y_k$, because $f$ is monotone increasing. So $f(x) leq y_k$.
If $x > a_k-1$, then there can be no such $x$ with $f(x) leq y_k-1$ because that would contradict the supremum property of $a_k-1$. So we have $y_k-1 < f(x)$.
So $$x in (a_k-1, a_k) Rightarrow f(x) in ( y_k-1, y_k]$$
calculus proof-verification
asked Aug 6 at 9:07
PhysicsUndergraduateStudent
527
asked Aug 6 at 9:07
PhysicsUndergraduateStudent
527
asked Aug 6 at 9:07
PhysicsUndergraduateStudent
527
asked Aug 6 at 9:07
PhysicsUndergraduateStudent
527
527
i don't see how $f(x) le y_k-1$ contradicts the supremum property of $a_k-1$. in fact the proof is much easier than you might think.
– pointguard0
Aug 6 at 9:28
@pointguard0 Because of the definition of $a_k-1$, it is the "largest" $x$ for which we have $f(x) leq y_k-1$. Any $x$ greater than $a_k-1$ does no longer suffice the inequality $f(x) leq y_k-1$. Or am I missing something?
– PhysicsUndergraduateStudent
Aug 8 at 13:46
and hence, $f(x) > y_k-1$ for all $x > a_k-1$. this is as simple as that
– pointguard0
Aug 8 at 14:06
@pointguard0 Yes, that's exactly what I wrote in my original post.
– PhysicsUndergraduateStudent
Aug 8 at 14:08
okay then, sorry, if i misunderstood. upvoted :)
– pointguard0
Aug 8 at 14:09
add a comment |Â
i don't see how $f(x) le y_k-1$ contradicts the supremum property of $a_k-1$. in fact the proof is much easier than you might think.
– pointguard0
Aug 6 at 9:28
@pointguard0 Because of the definition of $a_k-1$, it is the "largest" $x$ for which we have $f(x) leq y_k-1$. Any $x$ greater than $a_k-1$ does no longer suffice the inequality $f(x) leq y_k-1$. Or am I missing something?
– PhysicsUndergraduateStudent
Aug 8 at 13:46
and hence, $f(x) > y_k-1$ for all $x > a_k-1$. this is as simple as that
– pointguard0
Aug 8 at 14:06
@pointguard0 Yes, that's exactly what I wrote in my original post.
– PhysicsUndergraduateStudent
Aug 8 at 14:08
okay then, sorry, if i misunderstood. upvoted :)
– pointguard0
Aug 8 at 14:09
i don't see how $f(x) le y_k-1$ contradicts the supremum property of $a_k-1$. in fact the proof is much easier than you might think.
– pointguard0
Aug 6 at 9:28
i don't see how $f(x) le y_k-1$ contradicts the supremum property of $a_k-1$. in fact the proof is much easier than you might think.
– pointguard0
Aug 6 at 9:28
@pointguard0 Because of the definition of $a_k-1$, it is the "largest" $x$ for which we have $f(x) leq y_k-1$. Any $x$ greater than $a_k-1$ does no longer suffice the inequality $f(x) leq y_k-1$. Or am I missing something?
– PhysicsUndergraduateStudent
Aug 8 at 13:46
@pointguard0 Because of the definition of $a_k-1$, it is the "largest" $x$ for which we have $f(x) leq y_k-1$. Any $x$ greater than $a_k-1$ does no longer suffice the inequality $f(x) leq y_k-1$. Or am I missing something?
– PhysicsUndergraduateStudent
Aug 8 at 13:46
and hence, $f(x) > y_k-1$ for all $x > a_k-1$. this is as simple as that
– pointguard0
Aug 8 at 14:06
and hence, $f(x) > y_k-1$ for all $x > a_k-1$. this is as simple as that
– pointguard0
Aug 8 at 14:06
@pointguard0 Yes, that's exactly what I wrote in my original post.
– PhysicsUndergraduateStudent
Aug 8 at 14:08
@pointguard0 Yes, that's exactly what I wrote in my original post.
– PhysicsUndergraduateStudent
Aug 8 at 14:08
okay then, sorry, if i misunderstood. upvoted :)
– pointguard0
Aug 8 at 14:09
okay then, sorry, if i misunderstood. upvoted :)
– pointguard0
Aug 8 at 14:09
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
You already wrote
If $x < a_k$, then $f(x) leq f(a_k)$
Now, in the sentence above, make the following replacements:
- Replace $x$ with $a_k-1$
- Replace $a_k$ with $x$
What inequality do you get?
answered Aug 6 at 9:43
5xum
81.8k382146
$f(a_k-1) leq f(x)$, now what?
– PhysicsUndergraduateStudent
Aug 8 at 13:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You already wrote
If $x < a_k$, then $f(x) leq f(a_k)$
Now, in the sentence above, make the following replacements:
- Replace $x$ with $a_k-1$
- Replace $a_k$ with $x$
What inequality do you get?
answered Aug 6 at 9:43
5xum
81.8k382146
$f(a_k-1) leq f(x)$, now what?
– PhysicsUndergraduateStudent
Aug 8 at 13:39
add a comment |Â
up vote
0
down vote
You already wrote
If $x < a_k$, then $f(x) leq f(a_k)$
Now, in the sentence above, make the following replacements:
- Replace $x$ with $a_k-1$
- Replace $a_k$ with $x$
What inequality do you get?
answered Aug 6 at 9:43
5xum
81.8k382146
$f(a_k-1) leq f(x)$, now what?
– PhysicsUndergraduateStudent
Aug 8 at 13:39
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You already wrote
If $x < a_k$, then $f(x) leq f(a_k)$
Now, in the sentence above, make the following replacements:
- Replace $x$ with $a_k-1$
- Replace $a_k$ with $x$
What inequality do you get?
answered Aug 6 at 9:43
5xum
81.8k382146
You already wrote
If $x < a_k$, then $f(x) leq f(a_k)$
Now, in the sentence above, make the following replacements:
- Replace $x$ with $a_k-1$
- Replace $a_k$ with $x$
What inequality do you get?
answered Aug 6 at 9:43
5xum
81.8k382146
answered Aug 6 at 9:43
5xum
81.8k382146
answered Aug 6 at 9:43
5xum
81.8k382146
answered Aug 6 at 9:43
5xum
81.8k382146
81.8k382146
$f(a_k-1) leq f(x)$, now what?
– PhysicsUndergraduateStudent
Aug 8 at 13:39
add a comment |Â
$f(a_k-1) leq f(x)$, now what?
– PhysicsUndergraduateStudent
Aug 8 at 13:39
$f(a_k-1) leq f(x)$, now what?
– PhysicsUndergraduateStudent
Aug 8 at 13:39
$f(a_k-1) leq f(x)$, now what?
– PhysicsUndergraduateStudent
Aug 8 at 13:39
add a comment |Â
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i don't see how $f(x) le y_k-1$ contradicts the supremum property of $a_k-1$. in fact the proof is much easier than you might think.
– pointguard0
Aug 6 at 9:28
@pointguard0 Because of the definition of $a_k-1$, it is the "largest" $x$ for which we have $f(x) leq y_k-1$. Any $x$ greater than $a_k-1$ does no longer suffice the inequality $f(x) leq y_k-1$. Or am I missing something?
– PhysicsUndergraduateStudent
Aug 8 at 13:46
and hence, $f(x) > y_k-1$ for all $x > a_k-1$. this is as simple as that
– pointguard0
Aug 8 at 14:06
@pointguard0 Yes, that's exactly what I wrote in my original post.
– PhysicsUndergraduateStudent
Aug 8 at 14:08
okay then, sorry, if i misunderstood. upvoted :)
– pointguard0
Aug 8 at 14:09