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Want to show: $f$ is Lebesgue integrable and the value of the Lebesgue integral is the same is the Riemann integral. (We're not supposing that these two are equal when the Riemann integral exists).

I know (can prove) that this function will be both Riemann and Lebesgue integrable since $f$ is monotone and $[a,b]$ is measurable.

Here's what I have:

For any $\varepsilon>0$ , I want to show $|(A-a)-(B-b)|<\varepsilon$ where $A,a$ are the infimum and supremum of the Lebesgue lower and upper sums, respectively; and $B,b$ are the infimum and supremum of the lower and upper Riemann sums. Let $\varepsilon>0$. *Since $f$ is both Riemann and Lebesgue integrable, I can find partitions $P, Q$ of $[a,b]$ for which $|A-a|<\varepsilon/2$ and $|B-b|<\varepsilon /2$. Using the triangle inequality we have $|(A-a)-(B-b)|<\varepsilon$

Concerns:

*Am I allowed to consider $Q$, the partition with respect to the Lebesgue integral, as a partition of $[a,b]$ rather than $[f(a), f(b)]$?

**Doesn't the partition with respect to the Lebesgue integral start along the x-axis and then goes to the y-axis by taking inverse images of sets of the form $[a_i, a_{i+1}]$?

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3 Answers 3

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As Martin mentioned, $$\text{Riemann Integrable}\implies \text{Lebesgue Integrable}$$

THEOREM Let $f:[a,b]\to\Bbb R$ be a monotone function. We show that $f$ is Riemann integrable.

PROOF Let $P=\{t_0,\dots,t_n\}$ be a partition of $[a,b]$. We set the upper and lower sums:

$$U(f,P)=\sum_{k=1}^n M_k (t_k-t_{k-1})$$

$$L(f,P)=\sum_{k=1}^n m_k (t_k-t_{k-1})$$

where $$m_k=\inf\limits_{[t_{k-1},t_k]}f(x)$$

$$M_k=\sup\limits_{[t_{k-1},t_k]}f(x)$$

Assume $f$ increasing. Then $f$ is automatically bounded $f(a)\leq f(x)\leq f(b)$.

$$m_k=\inf\limits_{[t_{k-1},t_k]}f(x)=f(t_{k-1})$$

$$M_k=\sup\limits_{[t_{k-1},t_k]}f(x)=f(t_k)$$

This means that, for any partition $P$, we will have

$$U(f,P)-L(f,p)=\sum_{k=1}^n (f(t_k)-f(t_{k-1}))(t_k-t_{k-1})$$

Now, choose the partition $P$ such that $t_k-t_{k-1}<\delta$. Then $$\displaylines{ U(f,P) - L(f,p) = \sum\limits_{k = 1}^n {(f(} {t_k}) - f({t_{k - 1}}))({t_k} - {t_{k - 1}}) \cr < \delta \sum\limits_{k = 1}^n {(f(} {t_k}) - f({t_{k - 1}})) \cr < \delta \left( {f\left( b \right) - f\left( a \right)} \right) \cr} $$ Thus, it suffices to take $$\delta = \frac{\varepsilon }{{f\left( b \right) - f\left( a \right)}}$$ and we will have $$U(f,P) - L(f,p) < \varepsilon $$ whence $f$ will be (Riemann) integrable. For $f$ nonincreasing, apply the result to $-f$; which is non decreasing.

Another interesting fact is

THEOREM Let $f:[a,b]\to\Bbb R$ be monotone. Then the set $$\Delta=\{x\in[a,b]:f \text{ is discontinuous at } x\}$$ is at most countable.

PROOF

Define the function $$s\left( x \right) = \mathop {\lim }\limits_{y \to {x^ + }} f\left( y \right) - \mathop {\lim }\limits_{y \to {x^ - }} f\left( y \right)$$

since $f$ is monotone the left and right handed limits will always exist. It is readily seen that $f$ is discontinuous at $x=a$ if and only if $s(a)>0$. Note that, for any $x_1,x_2,\dots,x_n\in[a,b]$, we have $$\tag 1 0\leq \sum_{k=1}^n s(x_k)\leq f(b)-f(a)$$ (the sum of the gaps can't be greater than the whole $f(b)-f(a)$ gap). Let $L>0$ be given, and consider the set $$\Delta_L=\{x\in[a,b]:s(x)>L\}$$ We show this set is finite for each $L$. Indeed, suppose there were infinitely many points in $\Delta_L$. Then, we'd have $$\sum\limits_{k = 1}^n {s\left( {{x_k}} \right)} > \sum\limits_{k = 1}^n L = nL$$ and we could make this greater than $f(b)-f(a)$ by choosing $n$ large enough, which we can for we have infintely many points to take. But this contradicts $(1)$. Now, consider $\Delta_{1/n}$, for $n\in \Bbb N$. We know that this set is finite for each $n$, so $$\bigcup\limits_{n = 1}^\infty {{\Delta _{1/n}}} $$ is at most countable. But

$$\bigcup\limits_{n = 1}^\infty {{\Delta _{1/n}}} = \Delta $$

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I'm not supposed to assume that Riemann integrable implies Lebesgue integrable. That's why I introduced two partitions in my attempted proof. –  cap Nov 4 '12 at 23:43
    
@cap Can you try and prove that using the correspondence between step functions and indicator functions over intervals, and the fact $f$ is R.I. $\iff$ there exist step functions such that $s\leq f\leq t$ and $\int_a^b t-\int_a^b s<\epsilon$? –  Pedro Tamaroff Nov 4 '12 at 23:46
    
See this –  Pedro Tamaroff Nov 4 '12 at 23:48
    
We haven't discussed simple functions. I do know that RI implies LI, but I was hoping to show that they were the same here without using that fact. Is there a hole in my attempted proof? Thanks for the link and the input. –  cap Nov 4 '12 at 23:52
    
@cap: if you don't use simple functions, what is your definition of Lebesgue integral? I'm not saying it can't be done in a different way, but the standard definition is through simple functions. –  Martin Argerami Nov 5 '12 at 1:35
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Theorem 3.27 here gives a proof of this for $[0,1]$, it should be clear how to generalise that. It explains the partition used in a fair bit of detail, so hopefully this will help!

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And Martin is right; the correct statement is that Riemann integrable $\Rightarrow$ Lebesgue integrable, and the values of the integrals agree. –  user123123 Nov 4 '12 at 22:59
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I think it goes the opposite way as you see it. If a function is Riemann integrable, then it is Lebesgue integrable and the Lebesgue integral agrees with the Riemann integral (this last thing is easy to see, because the Riemann sums can be seen as integrals of simple functions).

You say you can prove that a monotone function is Riemann integrable; then you are done.

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Yes, but in my post I said that I can't assume that "If a function is Riemann integrable, then it is Lebesgue integrable and the Lebesgue integral agrees with the Riemann integral" –  cap Nov 4 '12 at 23:01
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