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Why do these relations stand??

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hey!!! :)

Let $f,g: [a,b] \to \mathbb{R}$ integrable functions.Show that: $\int_{a}^{b}(f+g)=\int_{a}^{b}f+\int_{a}^{b}g$

We suppose the subdivision $P=\{a=t_0<t_1<.....<t_n=b\}$ of $[a,b]$.

Let $t \in [t_k,t_{k+1}]$.

$$f(t) \leq sup f([t_k,t_{k+1}])$$
$$f(t) \geq inf f([t_k,t_{k+1}])$$
$$g(t) \leq sup g([t_k,t_{k+1}])$$
$$g(t) \geq inf g([t_k,t_{k+1}])$$

From these relations we get: $$u(f+g,P) \leq u(f,P)+u(g,P)$$
$$L(f+g,P) \geq L(f,P)+L(g,P)$$

where $L$ the lower sum and $U$ the upper sum.

As $f$ is integrable, $\forall \epsilon'>0 \exists P_1$ of $[a,b]$ such that $u(f,P_1)-L(f,P_1)<\epsilon'$.

As $g$ is integrable, $\forall \epsilon'>0 \exists P_2$ of $[a,b]$ such that $u(g,P_2)-L(g,P_2)<\epsilon'$.

We pick $P=P_1 \cup P_2$ and we get: $\int_{a}^{b}f+\int_{a}^{b}g< \epsilon + \underline{\int_{a}^{b}}(f+g)$
But why from this relation do we get: $\int_{a}^{b}f+\int_{a}^{b}g \leq \underline{\int_{a}^{b}}(f+g)$ ??

We also get: $\int_{a}^{b}f+\int_{a}^{b}g \geq \underline{\int_{a}^{b}}(f+g)- \epsilon$.And from this realtion,how do we get: $\int_{a}^{b}f+\int_{a}^{b}g \geq \underline{\int_{a}^{b}}(f+g)$ ? (Thinking)
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
As $f$ is integrable, $\forall \epsilon'>0 \exists P_1$ of $[a,b]$ such that $u(f,P_1)-L(f,P_1)<\epsilon'$.

As $g$ is integrable, $\forall \epsilon'>0 \exists P_2$ of $[a,b]$ such that $u(g,P_2)-L(g,P_2)<\epsilon'$.

We pick $P=P_1 \cup P_2$ and we get: $\int_{a}^{b}f+\int_{a}^{b}g< \epsilon + \underline{\int_{a}^{b}}(f+g)$
But why from this relation do we get: $\int_{a}^{b}f+\int_{a}^{b}g \leq \underline{\int_{a}^{b}}(f+g)$ ??
Because if $x,y\in\Bbb R$, then
\[
(\forall\epsilon>0\;x<y+\epsilon)\iff x\le y
\]
It's just a property of real numbers (of course, provable from axioms and the definitions of $<$ and $\le$).
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Let me expand a bit on Evgeny's answer:

The statement:

$(\forall \epsilon > 0: x < y + \epsilon) \iff x \leq y$

is "uinversally quantified" on the left, in other words it depends on it being true for EVERY such $\epsilon$.

The negation of such a statement is "existentially quantified" not a "universal negation".

In other words, the negation of:

$(\forall \epsilon > 0: x < y + \epsilon)$

is NOT:

$(\forall \epsilon > 0: x \geq y + \epsilon)$

but rather:

$(\exists \epsilon > 0: x \geq y + \epsilon)$

The reason I mention this, is that it is easier to prove the equivalence of the negation of Evgeny's statements rather than the originals, since we only have to deal with one particular $\epsilon$, the "counter-example" (this is called proof by contra-positives).

So we will show:

$(\exists \epsilon > 0: x \geq y + \epsilon) \iff x > y$

which is easier to do:

Suppose $x > y$: Then $x - y > 0$, and we can choose $\epsilon = x - y$, from which it is immediate that:

$x = x + (y - y) = (x + y) - y = (y + x) - y = y + (x - y) = y + \epsilon$.

Since $x \geq x$, we have the desired result.

On the other hand, suppose we just have some $\epsilon$ for which:

$x \geq y + \epsilon$.

Then:

$x \geq y + \epsilon > y$ (since $\epsilon > 0$).
 

solakis

Active member
Dec 9, 2012
304
the equivalence of the negation of Evgeny's statements rather than the originals, since we only have to deal with one particular $\epsilon$, the "counter-example" (this is called proof by contra-positives).

So we will show:

$(\exists \epsilon > 0: x \geq y + \epsilon) \iff x > y$
How can you prove that the negation of Evgeny's statement is:


$(\exists \epsilon > 0: x \geq y + \epsilon) \iff x > y$
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
This follows by purely logical reasoning using the facts that $(A\Leftrightarrow B)\iff (\neg A\Leftrightarrow\neg B)$, $(\neg x\le y)\iff x>y$ and $(\neg x< y)\iff x\ge y$.
 

solakis

Active member
Dec 9, 2012
304
This follows by purely logical reasoning using the facts that $(A\Leftrightarrow B)\iff (\neg A\Leftrightarrow\neg B)$, $(\neg x\le y)\iff x>y$ and $(\neg x< y)\iff x\ge y$.
Deveno claims that : ~(A<=>B) IT IS equivalent to ~A<=> ~B ,


The reason I mention this, is that it is easier to prove the equivalence of the negation of Evgeny's statements rather than the originals,

So we will show:

$(\exists \epsilon > 0: x \geq y + \epsilon) \iff x > y$
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Deveno claims that : ~(A<=>B) IT IS equivalent to ~A<=> ~B ,
I don't think so.

My version:
\[
(\forall\epsilon>0\;x<y+\epsilon)\iff x\le y
\]
Deveno's version:
So we will show:

$(\exists \epsilon > 0: x \geq y + \epsilon) \iff x > y$
Here $A$ is $\forall\epsilon>0\;x<y+\epsilon$; $\neg A$ is $\exists \epsilon > 0\; x \geq y + \epsilon$; $B$ is $x\le y$; $\neg B$ is $x > y$. In these notations, my version is $A\Leftrightarrow B$ and Deveno's is $\neg A\Leftrightarrow\neg B$.
 

solakis

Active member
Dec 9, 2012
304
I don't think so.

My version:


Deveno's version:


Here $A$ is $\forall\epsilon>0\;x<y+\epsilon$; $\neg A$ is $\exists \epsilon > 0\; x \geq y + \epsilon$; $B$ is $x\le y$; $\neg B$ is $x > y$. In these notations, my version is $A\Leftrightarrow B$ and Deveno's is $\neg A\Leftrightarrow\neg B$.

By what law is the negation of:

$\forall\epsilon>0\;x<y+\epsilon$
$\exists \epsilon > 0\; x \geq y + \epsilon$;
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Let me expand a bit on Evgeny's answer:

The statement:

$(\forall \epsilon > 0: x < y + \epsilon) \iff x \leq y$

is "uinversally quantified" on the left, in other words it depends on it being true for EVERY such $\epsilon$.

The negation of such a statement is "existentially quantified" not a "universal negation".

In other words, the negation of:

$(\forall \epsilon > 0: x < y + \epsilon)$

is NOT:

$(\forall \epsilon > 0: x \geq y + \epsilon)$

but rather:

$(\exists \epsilon > 0: x \geq y + \epsilon)$

The reason I mention this, is that it is easier to prove the equivalence of the negation of Evgeny's statements rather than the originals, since we only have to deal with one particular $\epsilon$, the "counter-example" (this is called proof by contra-positives).

So we will show:

$(\exists \epsilon > 0: x \geq y + \epsilon) \iff x > y$

which is easier to do:

Suppose $x > y$: Then $x - y > 0$, and we can choose $\epsilon = x - y$, from which it is immediate that:

$x = x + (y - y) = (x + y) - y = (y + x) - y = y + (x - y) = y + \epsilon$.

Since $x \geq x$, we have the desired result.

On the other hand, suppose we just have some $\epsilon$ for which:

$x \geq y + \epsilon$.

Then:

$x \geq y + \epsilon > y$ (since $\epsilon > 0$).
Don't we know that it is true for each $\epsilon$???
As we use the definition that $f$ and $g$ are integrable like that:
As $f$ is integrable, $\forall \epsilon'>0 \exists P_1$ of $[a,b]$ such that $u(f,P_1)-L(f,P_1)<\epsilon'$.

As $g$ is integrable, $\forall \epsilon'>0 \exists P_2$ of $[a,b]$ such that $u(g,P_2)-L(g,P_2)<\epsilon'$

and then we pick $\epsilon=\frac{\epsilon'}{2}$.Or am I wrong?? :confused:
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
"Naive" explanation:

"Not true for all" means "at least one exception".
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
By what law is the negation of:

$\forall\epsilon>0\;x<y+\epsilon$
$\exists \epsilon > 0\; x \geq y + \epsilon$;
Deveno is correct. For more details see Wikipedia.

Don't we know that it is true for each $\epsilon$???
Please don't overquote. It's hard to know what "it" means after a quote of a page of text ("it is true for each $\epsilon$").
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Don't we know that it is true for each $\epsilon$???
As we use the definition that $f$ and $g$ are integrable like that:
As $f$ is integrable, $\forall \epsilon'>0 \exists P_1$ of $[a,b]$ such that $u(f,P_1)-L(f,P_1)<\epsilon'$.

As $g$ is integrable, $\forall \epsilon'>0 \exists P_2$ of $[a,b]$ such that $u(g,P_2)-L(g,P_2)<\epsilon'$

and then we pick $\epsilon=\frac{\epsilon'}{2}$.Or am I wrong?? :confused:
Yes, the definition of "integrable" means the upper sums and lower sums can be made as close to each other as we like (this is what we MEAN by invoking "epsilon").

So if we want the upper and lower sums of the function $f+g$ to differ by less than epsilon, we can use the integrability of $f$ and $g$ to find two partitions of $[a,b]$ with each difference of lower and upper sums of $f$ and $g$ less than $\epsilon/2$.

We then REFINE those two partitions to a common partition (this is what taking the union does: chops up $[a,b]$ into more pieces, so that "clumps" of P can be formed to make the same pieces of $P_1$ and $P_2$ we used for our upper and lower sums originally). This refinement might make the lower sums "bigger", and the upper sums "smaller", so the difference between the two may decrease, but does not grow. This gives us:

$U(f+g,P) - L(f+g,P) \leq U(f,P) + U(g,P) - L(f,P) - L(g,P)$

$ = U(f,P) - L(f,P) + U(g,P) - L(g,P) \leq U(f,P_1) - L(f,P_1) + U(g,P_2) - L(g,P_2)$

$ < \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} = \epsilon$

Which simultaneously shows that:

\(\displaystyle \int_a^b f+g\) exists, and that:

\(\displaystyle \int_a^b f+g = \int_a^b f + \int_a^b g\).
 

solakis

Active member
Dec 9, 2012
304
Because if $x,y\in\Bbb R$, then
\[
(\forall\epsilon>0\;x<y+\epsilon)\iff x\le y
\]
It's just a property of real numbers (of course, provable from axioms and the definitions of $<$ and $\le$).
The formula:

\[
(\forall\epsilon>0\;x<y+\epsilon)\iff x\le y
\] does not mean anything.

It is not a wff.

Check your rules of wff ,and if you find a rule justifying your formula ,;let me know.

However the following formulas:

$\forall\epsilon (\epsilon>0\wedge x<y+\epsilon)\Longleftrightarrow x\leq y$

$\forall\epsilon (\epsilon>0\vee x<y+\epsilon)\Longleftrightarrow x\leq y$

$\forall\epsilon (\epsilon>0\Longrightarrow x<y+\epsilon)\Longleftrightarrow x\leq y$

Are open wff and can be negated


The following is a closed wff.

$\forall x\forall y[\forall\epsilon (\epsilon>0\Longrightarrow x<y+\epsilon)\Longleftrightarrow x\leq y]$
Describing the theorem you had in mind

Deveno copied your formula .

And the funny thing is that you refered me to Wikipedia where none of the formulas there support your formula
 
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