# Homework Help: 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

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

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

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

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

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

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

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

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

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.

21. Nov 7, 2012

### Zondrina

I just want to show why f is integrable as it will make showing -f and 1/f incredibly easy to show as it will be a reproduction of the proof.

Now, we are given that f is integrable on [a,b]. We also know that sup(f) = M and inf(f) = m. Which saves us writing some latex stuff out.

So we get our lower and upper sums :

$s_p = \sum_{i=1}^{n} m Δx_i = m(b-a)$ and $S_p = \sum_{i=1}^{n} M Δx_i = M(b-a)$.

So we get I = sup(sp) = m(b-a) and J = inf(Sp) = M(b-a).

Hence we get that I = m ≤ M = J.

Hmm have I done something wrong here?

22. Nov 7, 2012

### micromass

Huh??? But you are given that f is integrable. Why do you need to prove it then??

23. Nov 7, 2012

### Zondrina

I'm simply going to re-produce the proof for -f and 1/f. Since I already did the work of finding bounds for -f and 1/f I figured if I wrote out the proof for f, then it's going to be another copy and pasta for both of them.

I'm just curious to know why the proof I wrote didn't come out as expected.

24. Nov 7, 2012

### micromass

The difference is that you are given that f is integrable. And you need to prove that 1/f and -f are integrable. You can't just prove it for f (a proof which is just the hypothesis) and say that 1/f and -f are analogous.

25. Nov 7, 2012

### Zondrina

I suppose I got a bit lazy there then. I'll write out the proofs formally then.

We want to show that -f is integrable on [a,b] and we know that -m = sup(-f) and -M = inf(-f). So we define our lower and upper sums :

$s_p = - \sum_{i=1}^{n} MΔx_i = -M(b-a)$ and $S_p = - \sum_{i=1}^{n} mΔx_i = -m(b-a)$.

Is this better?