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I'm trying to work my way through Bartle's Elements of Integration and I am currently in Chapter 4, entitled the integral. If $f$ is a nonnegative measurable function, then its integral is defined as

$$\int f\,d\mu=\text{sup}\left\{\int \phi\,d\mu:\ 0\le\phi\le f\right\},$$

where $\phi$ is a simple measurable function. I've managed to prove the monotone convergence theorem, Fatou's lemma, and a few corollaries. However, at this point in time, there has not been any connection between these integrals and the Riemann integrals.

Then problem 4.L appears at the end of the chapter.

4.L. Let $X$ be a finite closed interval $[a,b]$ in $R$, let $\mathcal{X}$ be the collection of Borel sets in $X$, and let $\lambda$ be the Lebesgue measure on $\mathcal{X}$. If $f$ is a nonnegative continuous function on $X$, show that

$$\int f\,d\lambda=\int_a^b f(x)\,dx,$$

where the right side denotes the Riemann integral of $f$. (Hint: First establish this equality for a nonnegative step function, that is, a linear combination of characteristic functions of intervals.)

I'd appreciate any help I can receive with this question as it would open a door for me and help with my progress.

Thanks.

David
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  • The given hint is pretty good. Then you know you can approximate the Riemann integral by Riemann sums, and that takes you there. – Daniel Fischer Jul 02 '13 at 21:25
  • An even stronger hint: look at the proof of Lemma 2.11. This is a fundamental way of approximating non-negative measurable functions by simple functions. – gogurt Jul 02 '13 at 22:16

1 Answers1

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The first place to start is with their hint: we know that we can approximate Lebesgue integrals by integrals of simple functions, and Riemann integrals by Riemann sums; so, we stand a good chance if we can equate integrals of simple functions and Riemann integrals in some way.

If we use the "usual" notation for Riemann sums: break $[a,b]$ up in to $n$ even parts, each of length $\Delta x$, with endpoints $x_0=a,x_1,\ldots,x_n=b$, and with a specified chosen point $x_i^*\in[x_{i-1},x_i)$, then $$ \sum_{i=1}^{n}f(x_i^*)\,\Delta x=\int\left(\sum_{i=1}^{n}f(x_i^*)\chi_{[x_{i-1},x_i)}\right)\,d\lambda, $$ where $\chi_A$ is the characteristic function of the set $A$. This is immediate: $x_i^*$ is constant for each $i$, and $$ \int\chi_{[x_{i-1},x_i)}\,d\lambda=\lambda([x_{i-1},x_i))=x_i-x_{i-1}=\Delta x. $$ (Of course, we can do this with uneven partitions too... but no need!)

Now, of course the Lebesgue integral of a non-negative function $f$ is defined in terms of simple functions $\phi$ such that $\phi\leq f$; so, you need to come up with specific characteristic functions which are always lower bounds on $f$.

Hint: these might correspond to a special Riemann sum that is always a lower bound on $\int_a^b f(x)\,dx$.

From here, there are lots of ways to go. Two possible ways to proceed, depending on your preference:

1) Suppose that at each step, you get a new Riemann sum / simple function by splitting each of your previous intervals in half. Does the "special Riemann sum" from the hint behave nicely? The Monotone Convergence Theorem might come in someplace.

2) If you prefer to be less sneaky, you could also do this: for any $\epsilon>0$, can you prove that there is a simple function $\phi$, $\phi\leq f$, such that $\int\phi\,d\lambda$ differs from $\int f\,d\lambda$ by at most $\epsilon$ and the associated Riemann sum is within $\epsilon$ of $\int_a^b f(x)\,dx$?

Nick Peterson
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