# A Step Function Proof

1. Oct 10, 2007

### brh2113

1. The problem statement, all variables and given/known data
Let $$f$$ be a function that is integrable on $$[a,b]$$ and bounded by
$$0$$ $$\leq f(x)\leq M$$ for some $$M$$. Prove that $$f^{2}$$ is also integrable on this interval.

2. Relevant equations
We've done many problems with step functions $$s(x)$$$$\leq f(x) \leq t(x)$$,

where $$s(x)$$ and $$t(x)$$ are step functions. Then we've defined a

function's lower integral as the supremum of the set of the values of all $$\int_a^b s(x)dx$$

provided that $$s(x)$$ $$\leq f(x)$$ and the function's upper integral as the

infimum of the set of all $$\int_a^bt(x)dx$$ provided that

$$f(x)$$ $$\leq t(x)$$ on $$[a,b]$$.

3. The attempt at a solution

Attached is my work. It's a little sloppy, so I'll explain what I tried here:

I first tried using the weighted mean value theorem, but all that that does is show that $$f^{2}$$ is bounded.

I then drew a line and tried again. However, I've realized that my definition of the step functions is faulty because the function is not monotonic, so the second attempt is wrong. As a result, I've been unable to represent my step functions more explicitly using summation notation and expanding it, so I haven't been able to work with the equations. There must be something I'm missing.

If the function were continuous, then I'd know that $$f^{2}$$ is continuous and I'd be done, but unfortunately it is not.

The only thing I know is that $$\int_a^b f(x)dx$$ is bounded and exists, and so $$\int_a^b f(x)^{2}dx$$ is also bounded.

Again, if I can show that the lower integral and the upper integral are equal, then by our definition I will have proven that the function is integrable.

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2. Oct 10, 2007

### EnumaElish

Since f is positive, f2 is monotonic. So s < f < t implies s2 < f2 < t2.

3. Oct 10, 2007

### brh2113

Unfortunately f is not monotonic, since it is not continuous. For example, f(a) could equal 5, and f(b) = 3, when a<b, provided that 3 and 5 are less than M. Moreover, f^2 is neither continuous nor monotonic, but I think I can assert that inequality.

Still, it doesn't seem to get me anywhere. Because the function is not monotonic, I cannot write my summation notation explicitly enough that it can be expanded and canceled to show that the difference between the supremum of s and the infimum of t is 0.

I appreciate the help, though. I spent about an hour today working with a T.A. with summation notation, but we ended up getting nowhere. It's possible that that's necessary, but I'm wondering if there's a simpler way to do it.

4. Oct 11, 2007

### EnumaElish

I meant that f^2 is monotonic with respect to f, or "square" operation is a monotone transformation. I did not realize that you couldn't represent f as a step function so I was taking it as a given.

You need to think what the term "integrable" exactly means.

Last edited: Oct 11, 2007
5. Oct 11, 2007

### brh2113

Oh monotone transformation. I'm not familiar with that term, but I think I understand what you're saying to mean that each value of f^2 corresponds to one value of f?

Anyways, I do agree with that inequality, and with the help of a TA in one of my classes I managed to work out the problem. Thanks again for the help; it's much appreciated.

It says in the instructions that I should edit the topic using the Thread Tools to say "Solved," but I can't see any option under thread tools to change the name. Can someone tell me through a message how to do this?