# Proving Riemann Integration

1. Nov 7, 2012

### Zondrina

1. The problem statement, all variables and given/known data

Suppose that f(x) is a bounded function on [a,b]

If M = sup(f) and m = inf(f), prove that -M = inf(-f) and -m = sup(-f). If m>0, show that 1/f is bounded and has 1/m as its supremum and 1/M as its infimum.

Let f be integrable on [a,b]. Prove that -f is integrable on [a,b] and if m>0, prove that 1/f is integrable on [a,b].

2. Relevant equations

So M is the least upper bound for f on [a,b] and m is the greatest lower bound for f on [a,b].

3. The attempt at a solution

So there's lots of stuff involved in this question. I'll start by trying to prove the first thing :

If m = inf(f) and M = sup(f), prove that -m = sup(-f) and -M = inf(-f).

So we know that : m ≤ f ≤ M

Forgive me if I'm wrong, but this seems like a one liner? Simply multiplying by (-1) gives :

-m ≥ -f ≥ -M so that -m is the least upper bound for -f and -M is the greatest lower bound for -f.

Before I go any further I'd like to check if I'm not jumping the gun a bit there.

Last edited: Nov 7, 2012
2. Nov 7, 2012

### micromass

Staff Emeritus
Yes, I think you are jumping the gun. You correctly deduced that

$$-M\leq -f\leq -m$$

and this shows that -m is an upper bound of -f. But why is it a least upper bound?? How does that follow from the above inequality? Same for -M being the greatest lower bound.

3. Nov 7, 2012

### Zondrina

Whoops I mistyped a few things in my original post. I fixed them now. I meant for M to be the sup and m to be the inf.

4. Nov 7, 2012

### micromass

Staff Emeritus
Please don't edit your original posts, it destroys the flow of the topic and is hard to read

Anyway, my reply still holds, you have only proven that -m is an upper bound, but why is it the least upper bound? And why is -M the greatest lower bound?

5. Nov 7, 2012

### Zondrina

Now as for what you were saying.

Since f is a bounded function on a nonempty set of real numbers, it has to have a least upper bound ( also a greatest lower bound ).

So if L1 is any upper bound for -f, and L2 is any lower bound for -f, then :

$L_1 ≥ -m ≥ -f ≥ -M ≥ L_2$

6. Nov 7, 2012

### micromass

Staff Emeritus
How did you obtain that inequality?? And why does that imply anything?

You need to prove: if a is another lower bound of -f, then a≤-M. And same for m.

7. Nov 7, 2012

### Zondrina

Ah yes, so I want to show that -m is the least upper bound and -M is the greatest lower bound.

Right now we already have that -m is an upper bound for -f and -M is a lower bound for -f.

If L1 is any upper bound bound for -f, we want to show L1 ≥ -m. So :

L1 ≥ -f
L1 ≥ -m

This seems too obvious? Showing it for the lower bound is going to be exactly the same so I'll focus on this.

8. Nov 7, 2012

### micromass

Staff Emeritus
So you say: if $L_1\geq -f$, then $L_1\geq -m$?? Can you clarify this? That doesn't seem obvious to me.

9. Nov 7, 2012

### Zondrina

Whoops, I had a brain hiccup there, forgot to do this :

L1 ≥ -f
-L1 ≤ f

The rest of the algebra is obvious if what I'm thinking is correct. I believe I can sub m in for f here?

10. Nov 7, 2012

### micromass

Staff Emeritus
OK, but can you write this out with some explanations instead of only writing down the inequalities??

For example, what you could writes:
We want to prove that -m is the greatest upper bound of -f. So let L be another upper bound of -f, this means: L≥-f. Multiplying by -1 yields f≤-L. Now .....

Can you complete the above? Try to write using sentences instead of just symbols.

11. Nov 7, 2012

### Zondrina

Yessir, so.

Now ... since we are given that m ≤ f ≤ M and that m is the infimum of f, we know f attains the value m on the interval [a,b]. That is, f = m at some point. So :

-L1 ≤ m
L1 ≥ -m

Hence L1 is an upper bound for -f and -m is a least upper bound for -f.

12. Nov 7, 2012

### micromass

Staff Emeritus
Why should f attain the value m somewhere on the interval [a,b]??
If you're using the extreme value theorem, then think again: it only holds for continuous functions.

13. Nov 7, 2012

### Zondrina

I just re-read the theorem, you're right. I haven't used it in so long I had forgotten. Continuity would've made things nice :(.

So I need a different explanation, but the rest of what I said looks good.

Now I argued that L1 was an upper bound for -f, so it would be a lower bound for f. Since m is the greatest lower bound for f, we get :

-L1 ≤ m
L1 ≥ -m

Hence L1 is an upper bound for -f and -m is a least upper bound for -f.

14. Nov 7, 2012

### micromass

Staff Emeritus
OK, that is good! (although the "Hence L1 is an upper bound of -f" is unnecessary since that was the hypothesis)

15. Nov 7, 2012

### Zondrina

Sweet, okay. I'll write everything nice and clean with a good explanation here then.

So we know that m ≤ f ≤ M which implies that -m ≥ -f ≥ -M. What we want to show that -m is the supremum of -f and that -M is the infimum of -f. Right now we have that -m is an upper bound for -f and -M is an lower bound for -f.

So suppose that L1 is any upper bound for -f. So we get L1 ≥ -f which yields -L1 ≤ f. Now, since we argued that L1 is any upper bound for -f, so it must be a lower bound for f. Since m is the greatest lower bound for f, we get -L1 ≤ m which implies that L1 ≥ -m. Hence -m is the least upper bound for -f.

Suppose now that L2 is any lower bound for -f. So we get L2 ≤ -f which yields -L2 ≥ f. Now, since we argued that L2 is any lower bound for -f, it must be an upper bound for f. Since M is the least upper bound for f, we get -L2 ≥ M which implies that L2 ≤ -M. Hence -M is the greatest lower bound for -f.

16. Nov 7, 2012

### Zondrina

Now I'm going to attempt the second part of the question.

If m>0, show that 1/f is bounded and has 1/m as its supremum and 1/M as its infimum.

So we know that m ≤ f ≤ M which implies that 1/m ≥ 1/f ≥ 1/M.

This looks like an exact copy and paste of my post above except with a few different numbers if I'm not mistaken?

17. Nov 7, 2012

### micromass

Staff Emeritus
OK, so that proves that 1/m is an upper bound of 1/f and that 1/M is a lower bound. So that implies indeed that 1/f is bounded. (where did you use that m>0 anyway?)

But you still have more work to do if you wand to show that 1/M is the greatest lower bound and that 1/m is the least upper bound. The work is very analogous to above though.

18. Nov 7, 2012

### Zondrina

I'm assuming that M can't be zero either otherwise we would have a problem here, but here goes my attempt :

So we know that m ≤ f ≤ M which implies that 1/m ≥ 1/f ≥ 1/M hence 1/f is bounded above by 1/m and bounded below by 1/M. We want to show that 1/m is the supremum of 1/f and 1/M is the infimum of 1/f. Right now we have that 1/m is an upper bound for 1/f and 1/M is a lower bound for 1/f.

So suppose that Q1 is any upper bound for 1/f. Then we have Q1 ≥ 1/f which implies that 1/Q1 ≤ f. Now since Q1 is any upper bound for 1/f, 1/Q1 must be a lower bound for f. Now, since m is the greatest lower bound for f, we have that 1/Q1 ≤ m which yields Q1 ≥ 1/m. Hence 1/m is the least upper bound for 1/f.

Now suppose that Q2 is any lower bound for 1/f. Then we have Q2 ≤ 1/f which implies that 1/Q2 ≥ f. Now since Q2 is any lower bound for 1/f, 1/Q2 must be an upper bound for f. Now, since M is the least upper bound for f, we have that 1/Q2 ≥ M which yields Q2 ≤ 1/M. Hence 1/M is the greatest lower bound for 1/f.

I believe that should do it.

19. Nov 7, 2012

### micromass

Staff Emeritus
That's correct! And you have written that perfectly!

20. Nov 7, 2012

### Zondrina

Whoop :D, so I only have one part left here : Let f be integrable on [a,b]. Prove that -f is integrable on [a,b] and if m>0, prove that 1/f is integrable on [a,b].

So I'll be using some notation here, but I'll try to make it clear what is what.

I have a definition here : If I = sup(sp) = inf(Sp) = J, then f is integrable.

Now, $s_p = \sum_{i=1}^{n} m_i Δx_i$ and $S_p = \sum_{i=1}^{n} M_i Δx_i$.

To fill in the last few blanks, $m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}$ and $M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}$

I'll start writing my proof in the next post since this is pretty cluttered as is.