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Theorem:

If $f: [a,b] \to \Bbb R$ is Riemann-integrable, then $f$ is Lebesgue-integrable, and:

$$ \int_a^b f(x) dx = \int_{[a,b]} fd\lambda$$

Here is the proof: (assuming that we know that step functions are Lebesgue integrable and that their Riemann integral coincides with their Lebesgue one)

The Riemann-integrability of $f$ implies the existence of two sequences of step functions $(g_n) \uparrow$ and $(h_n) \downarrow$, such that $g_n \le f \le h_n$ for all $n$, and:

  • $\int_a^b (h_n(x) - g_n(x))dx \to 0$ $(\star)$

  • $\int_a^b f(x) dx = \lim \int_a^bg_n(x)dx$

It is easy to see that $(g_n)$ and $(h_n)$ converge pointwise to some $g$ and $h$ $\in L^1$ (respectively).

Letting $\gamma = |g_1| + |h_1|$, we get that $h_n - g_n \le \gamma$ for all $n$, and $\gamma \in L^1$ being the sum of two $L^1$ functions. Moreover, $h_n - g_n \to h - g$ is a sequence of $L^1$ functions converging to an $L^1$ function. Therefore, by LDCT, we get:

$$\int_{[a,b]} (h_n - g_n) d \lambda \to \int_{[a,b]} (h - g) d \lambda$$

Which implies that:

$$\int_{[a,b]} (h - g) d \lambda = 0$$

Hence $h - g = 0$ $\lambda-$a.e. since it is $\ge 0$, i.e. $h = g$ $\lambda-$a.e.

Now, $g \le f \le h$, hence $f = g$ $\lambda-$a.e.

But from this, we can only conclude that $f$ is measurable with respect to the Lebesgue $\sigma-$algebra on $[a,b]$, and not the Borel one.

Are we assuming the Lebesgue $\sigma-$algebra on $[a,b]$? If not, how can we prove that $f$ is measurable?

The rest of the proof becomes straightforward.

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    When you construct the Lebesgue integral, you assume that you are working with the Lebesgue $\sigma$-algebra on $[a,b]$. – Ángel Valencia Jan 21 '16 at 21:55
  • @AngelValencia: our professor did not assume that, instead, we always talk about measurability with respect to the Borel $\sigma-$algbera; this is the source of my confusion. Can you give me some reference where this is explicitly stated? I searched the internet in vain. –  Jan 21 '16 at 21:58
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    I don't remember about explicit sources (I think Bartle's book "Elements of Integration" is a good choice), but when you speak about Lebesgue integral, you consider Lebesgue-measurable functions, that is, functions $f:[a,b]\to\mathbb{R}$ such that $f^{-1}(A)$ is Lebesgue-measurable for all Borel-measurable subset $A$ of $[a,b]$. – Ángel Valencia Jan 21 '16 at 22:14
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    @ÁngelValencia: That's a common approach, but it's not universal: it's entirely possible to define the Lebesgue integral by only working with Borel-measurable functions. – Nate Eldredge Jan 21 '16 at 22:19

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A Riemann-integrable function is not necessarily Borel, see A Riemann integrable function $f$ on a bounded interval $[a, b]$ is measurable with respect to the Borel measure on $[a,b]$?

So you are going to have to assume you are working with the Lebesgue $\sigma$-algebra if you want to get a true theorem.

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Nate Eldredge
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  • Thank you. Do you have in mind a good reference in which there is an elaborative construction of the Lebesgue integral? –  Jan 21 '16 at 22:05
  • @AhmedHussein: Honestly, that describes every book on basic measure theory. Royden's book is nice and focuses on $\mathbb{R}$; Folland is another nice book but works on abstract measure spaces. – Nate Eldredge Jan 21 '16 at 22:06
  • Heh. If you'd asked me I would have said a Riemann integrable function was Borel. Clearly not, the example is very simple. Something new every day... – David C. Ullrich Jan 21 '16 at 22:40