# The Lebesgue and Riemann integrals of an increasing function over $[a,b]$ are the same.

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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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

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