Question

PROVE: Let f be riemann integrable on [a,b] Let \(\alpha , \beta , \gamma \) be arbitrary numbers in [a,b] (not necessarily ordered in any way) Prove that \[\int^\gamma_\alpha f=\int^\beta_\alpha f+ \int^\gamma_\beta\]

Answer 1

just a hint here, I don't see how it is different from the thm 5.2.2 from last time

Answer 2

consider the case with \(\beta>\gamma\)

Answer 3

right, but wouldn't there be 4 potential options?

Answer 4

yeah I guess so, you would have really 6.....

Answer 5

beta greater than gamma is just that thm from the last problem right?

Answer 6

post that link again. I thought it was nested.

Answer 7

nested?

Answer 8

http://www.jirka.org/ra/realanal.pdf page157

Answer 9

it says b<c in that theorem

Answer 10

oh i guess i meant the \(\gamma<\beta\) option

Answer 11

yes that would be taken care of by that theorem

Answer 12

said it backwards shoot

Answer 13

so you have some rule about \(\int_a^bf=-\int_b^af\)?

Answer 14

yea, we have that

Answer 15

I guess you will have six cases then? \(a\le b\le c\\ a\le c\le b\\ b\le a \le c\\ b \le c \le a\\ c\le a \le b\\ c \le b \le a\)

Answer 16

so we need cases \(\gamma<\alpha ; \beta<\alpha ; \beta>\alpha; \beta>\gamma\)

Answer 17

we already have \(\beta<\gamma\), but do I have to do all of them? like 6! solutions?

Answer 18

there must be some general way

Answer 19

how do you have \(\beta \le \gamma\)?

Answer 20

we can discredit equals because that equals zero...

Answer 21

how do you have \(\beta < \gamma\)?

Answer 22

that thm from before

Answer 23

5.2.2?

Answer 24

it does not say that.

Answer 25

why not? Didn't we prove that?

Answer 26

lol im confused. we did not prove theorem 5.2.2 it is in your book. you said that theorem 5.2.2 says that we must have beta < gamma, and then you say we proved it. I think I am missing something.

Answer 27

brb, getting off of bus and will get on pc when I get home. Should not take long.

Answer 28

ok thanks, I'm so confused on what I need to prove here

Answer 29

I guess I also don't see how this works when we have \[\int^\gamma_\alpha f=\int^\beta_\alpha f+ \int_\gamma^\beta\]

Answer 30

Maybe prove a lemma first and use it instead of working all the cases :
\[\int \limits_{\alpha}^{\beta}f = -\int\limits_{\beta}^{\alpha} f\]

Answer 31

right but how is that the same?

Answer 32

when we subtract that? ... ohhhh ok I'm a nitwit, ok, so I just need to prove that now. Let me try that

Answer 33

so conceptually I get it, we are subtracting the excess because we have modified our values, but can I get a hint on the technical proof?

Answer 34

proving lemma should be straightforward right ?

Answer 35

you're asking how to use it to argue that the integral splitting does not depend on the order ?

Answer 36

hmm ok... I'll try let me hit my head off the wall for a few minutes and I'll be back

Answer 37

i don't get what you're asking

Answer 38

what you consider straightforward, is not to me

Answer 39

okay its straightforward in calc1 atleast
idk how it changes in real analysis

Answer 40

so I'm gonna try and see what I'm missing

Answer 41

\[\int_a^b f = F(b)-F(a) = -\left[F(a) - F(b)\right] = -\int_b^a f \]

Answer 42

oh that lemma, yea that has already been proved, I thought you meant the subtracion of the integral to cut off the excess

Answer 43

yeah making that connection is tricky, one sec

Answer 44

but first, do we have 6 pieces or do we have more?

Answer 45

so like if we use \(\gamma<\alpha\) do we then have 4 sub-cases where \(\beta <\alpha ; \beta<\gamma; \beta>\gamma; \beta >\alpha\)

Answer 46

which means we have 24 cases??? That seems like alot

Answer 47

there should be a clever way to argue instead of listing out all the different cases

Answer 48

i see only 6 cases though

Answer 49

hmm

Answer 50

the ones listed by zzr

Answer 51

yea, but isn't there multiple subcases if you bound it?

Answer 52

ie the limits are gamma and alpha, so you have two possibilities there

Answer 53

those 6 cases exhaust all the subcases

Answer 54

either gamma is bigger or alpha is bigger, then there are 4 subcases

Answer 55

beta is bigger/less than alpha, beta is bigger/less than gamma

Answer 56

all four for both options on bounds

Answer 57

list ur cases based on the "middle value"

Case1 : when alpha is in between beta and gamma, you get two subcases
\[\beta \le \alpha \le \gamma \]
\[\gamma \le \alpha\le \beta\]

Answer 58

but the end values are not set either

Answer 59

yes you get two mose cases with two subcases each ?

Answer 60

*more

Answer 61

how am I getting 10...

Answer 62

im not sure, actually im bit lost...

Answer 63

think of it like this :

you have 3 points on real line

Answer 64

you can permute them in 3! = 6 ways

Answer 65

... I abhor analysis. Thank god next week is the last week of it....

Answer 66

I am thinking 2 options 4 suboptions

Answer 67

somehow your extra cases must be redundant

Answer 68

hmm ok,

Answer 69

ah I see the redundant ones now

Answer 70

Oh good, im still trying to prune the redundant ones

Answer 71

but still, 6 cases seems like a lot, shouldn't there be another way?

Answer 72

So you work all 6 cases by subtracting the excess where necesssary ?

Answer 73

yeah im fully sure there will be another nice way but im not getting any other ideas as of nw

Answer 74

yea... but I also don't see how to technically and rigorously prove that

Answer 75

@zzr0ck3r is still off

Answer 76

Conceptually I fully understand it

Answer 77

same here idk real analysis terminology

Answer 78

it's the technical..... ughh