# Any Simple Example of Lebesgue Integration?

I saw several conceptual explanations regarding Lebesgue Integration, but can I see few practical examples that require Lebesgue Integration? What I need is just a toy case of Lebesgue Integration.

For instance, how can I integrate $f(x)=x$ using Lebesgue Integration? I know it is $F(x)=\frac{1}{2}x^2$ since Riemann Integration is applicable, but do not know how to do this using Lebesgue Integration instead.

Also can I apply Lebesgue Integration to compute moments of distributions? For example, can I compute $\mathrm{E}[X]=\lambda^{-1}$ for a random variable $X$ that follows the exponential distribution with $\lambda$?

• It’s not clear what you’re after, but a standard example that may be relevant is the integral of the indicator function of the rationals. – Ittay Weiss Dec 7 '17 at 22:07
• Suppose that the random variable $U$ follows the uniform distribution on $[0,1]$. What's the probability of getting rational numbers $\mathbb{P}(U\in\mathbb{Q})$? – GNUSupporter 8964民主女神 地下教會 Dec 7 '17 at 22:08
• I saw the rational-irrational function example so many times and roughly understand that it is the same because $\aleph(\mathbb{R})=\aleph(\mathbb{R}\setminus\mathbb{Q})$. However, I want to numerically apply Lebesgue Integration with a toy case. For instance, $\int_0^1{x\mathrm{d}x}=\lim_{n\rightarrow\infty}{\sum_{i=1}^{n}{\frac{1}{n}f\left(\frac{i}{n}\right)}}=\lim_{n\rightarrow\infty}{\frac{n^2+n}{2n^2}}=\frac{1}{2}$. Sorry for such a poor question. – Junyong Kim Dec 7 '17 at 22:37
• @JunyongKim Not sure about that notation $\aleph (\mathbb R)...$ makes it seem like you're talking about cardinality. If this is the case, this is the wrong way to look at it. For instance $\mathbb R$ and $\mathbb R\setminus [0,1]$ have the same cardinality, but $\int_{\mathbb R} 1_{[0,1]} d\mu \ne 0.$ What's important is that the rationals are measure zero. (Note all countable sets have measure zero but the converse is not true, so while cardinality is useful here it is important to keep the concepts of countable and measure zero distinct.) – spaceisdarkgreen Dec 8 '17 at 5:09

When they both exist, Lebesgue and Riemann integration give the same thing. In particular, the fundamental theorem of calculus, substitution theorems, etc, are just as true for the Lebesgue integral as for the Riemann integral. So when you use substitution to compute the expected value of an exponential as $$E(X) = \int_0^\infty x\lambda e^{-\lambda x}dx = \frac{1}{\lambda}$$ it doesn't matter at all whether it's a Lebesgue integral or a Riemann integral.

Nonetheless, we can certainly do a toy computation of $\int_0^1 x^2 dx$ "the Lebesgue way". The basic idea, as I'm sure you've heard, is to partition the $y$ axis rather than the $x$ axis. So let's take an even-spaced partition $A^n_i = [\frac{i-1}{n},\frac{i}{n})$ of $[0,1).$ Then we can partition the $x$ axis into $$B_i^n = \{x\in[0,1) \mid f(x)\in A_i^n\}.$$ Here this partition is simple since $f$ is increasing. We have $B_i^n = [ \sqrt{\frac{i-1}{n}},\sqrt{\frac{i}{n}}).$ (But note that for a non-monotonic function the $B_i^n$ could be the union of several intervals, not just a single interval).

Then, we can write a sequence of simple functions (i.e. step functions) $$s^{lower}_n(x) = \sum_{i=1}^n \frac{i-1}{n} 1_{B_i^n}(x)$$ where $1_E(x)$ is the indicator function that is one if $x\in E$ and zero otherwise. So it's a function that is locally constant on the sets $B_i^n$ but takes different values between the different sets... it looks like a staircase here. Notice that $\frac{i-1}{n}$ is the lower limit of the y-axis partition interval $A_i^n,$ so that $s^{lower}_n < f.$

Now, we define that for a constant function $f(x)=c$, the integral of the function over some set $E$ is just $c\mu(E)$ where $\mu(E)$ is the measure of $E,$ and we extend it to simple functions in the obvious way: $$\int \sum_{i=1}^n a_i 1_{E_i} \;d\mu = \sum_{i=1}^n a_i \mu(E_i).$$

So we have $$\int_{[0,1]} s^{lower}_n \;d\mu = \sum_{i=1}^n \frac{i-1}{n}\mu(B_i^n) = \sum_{i=1}^n \frac{i-1}{n}\left(\sqrt{\frac{i}{n}}-\sqrt{\frac{i-1}{n}} \right).$$

Similarly, we can define a function $s_n^{upper}(x) >f$ by $$s_n^{upper}(x)=\sum_{i=1}^n \frac{i}{n} 1_{B_i^n}(x)$$ and compute $$\int_{[0,1]} s^{upper}_n \; d\mu= \sum_{i=1}^n \frac{i}{n}\left(\sqrt{\frac{i}{n}}-\sqrt{\frac{i-1}{n}} \right).$$

I'm not sure if these sums can be computed in closed form, but in any event, it is true that $$\lim_{n\to\infty} \int_{[0,1]} s^{upper}_n \; d\mu = \lim_{n\to\infty} \int_{[0,1]} s^{upper}_n \; d\mu = \frac{1}{3},$$ so since $s_n^{upper} > f > s_n^{lower}$ for all $x,$ the integral of $f(x) = x^2$ is squeezed to $1/3.$

(Formally, the Lebesgue integral of a non-negative function is defined as $$\int_{[0,1]} f(x)d\mu = \sup_{s\in\Sigma_{[0,1]}} \left\{\int s d\mu \mid s <f\right\}$$ where $\Sigma_{[0,1]}$ is the set of all simple functions on $[0,1].$ We can show that for any simple, $s < f,$ $\int s\;d\mu \le \int s_n^{upper} d\mu,$ so we indeed have $\int f\;d\mu \le \int s_n^{upper} d\mu,$ and analogously for $s^{lower}.$)

This might be a little unsatisfying for your purpose, cause we can't compute the sum in closed form before you take the limit, like you could in the Riemann case (or I can't, anyway). But this was due to the fact that I chose an evenly spaced partition of the y-axis for clarity. If I instead let $A^n_i = [\left(\frac{i-1}{n}\right)^2,\left(\frac{i}{n}\right)^2),$ then we get $B^n_i = [\frac{i-1}{n},\frac{i}{n})$ and going through the same motions we get $$s^{lower}_n(x) = \sum_{i=1}^n \left(\frac{i-1}{n}\right)^2 \mu(B_n^i) = \sum_{i=1}^n \left(\frac{i-1}{n}\right)^2 \frac{1}{n}$$ which is precisely the same thing you'd write down for a Riemann sum computation (and that I presume you know how to compute in closed form). This is no accident, of course... Any proper Riemann-integrable function is Lebesgue integrable and the integrals have the same value.

Hopefully my indulgence of your question doesn't obscure what I said in the first paragraph. The "point" of Lebesgue integration is not that it's a way to do standard integrals of calculus by some new method. It's that the definition of the integral is more theoretically powerful: it leads to more elegant formalism and cleaner results (like the dominated convergence theorem) that are very useful in harmonic/functional analysis and probability theory. It's also true, as said in the comments, that it allows you to integrate more functions, like the indicator function of the rationals. This isn't really that compelling of an advantage in and of itself (and note that if we extend to improper integrals there are things that are Riemann-integrable but not Lebesgue), but the argument that it is zero shows how nice of an ally measure theory can be... a function that was ugly and too cumbersome for Riemann integration to deal with is tamed quite easily by Lebesgue.

• Truly appreciate for this meticulous advice. My apology but I have a few more things to ask. (a) What is the meaning of "simple" here? (b) You mentioned that the partition $B$ is simple since $f$ is increasing, but how can I know whether one partition is simple or not? (c) Why $\mu\left(B_i^n\right)=\sqrt{\frac{i}{n}}-\sqrt{\frac{i-1}{n}}$? I can intuitively understand this, but how can I in general evaluate $\mu(E)$ if $E$ is something really strange? – Junyong Kim Dec 8 '17 at 18:27
• (d) I studied, for example, how to deal with $\lim_{x\rightarrow0^+}$ and $\lim_{x\rightarrow0^-}$, but can I algebraically deal with $\sup$ and $\inf$? I know these conceptually, but not rigorously. (e) I heard that Lebesgue Integration requires $f$ to be "measurable," but I cannot even intuitively understand why this is required. Is there any exception at this toy level? Sorry again for my poor understanding. – Junyong Kim Dec 8 '17 at 18:28
• (a) I used simple in two different ways. The first is technical: a simple function is one that is a finite linear combination of indicator functions, like $s_n^{lower}$ or upper. (b) The second is colloquial: When I said the partition $B$ is simple, I just meant that it isn't very complicated in that each partition element is just an interval. To see why the partition of the $x$ axis might be more complicated than this, draw a function that is neither increasing nor decreasing and figure out what $B_i^n = \{x\in[0,1) \mid f(x)\in A_i^n\}$ looks like for some partition $A$ of the y axis. – spaceisdarkgreen Dec 8 '17 at 23:42
• (c) That is the measure cause the measure of any interval is its length. To compute the lebesgue measure of a more complicated measurable set is, well, more complicated. You can see the definition here en.wikipedia.org/wiki/Lebesgue_measure . (d) Working directly with the sup definition directly is indeed a bit cumbersome (which is why I used a squeezing argument). Fortunately once you slog through a technical proof of one of the main theorems (monotone convergence or dominated convergence), things become easier. (Remember, our goal is not really to compute the values of integrals.) – spaceisdarkgreen Dec 8 '17 at 23:57
• (a) So a simple function is a finite linear combination of indicator functions, but then $s_n^{lower}$ would not be a simple function anymore as soon as one applies $\lim_{n\rightarrow\infty}$ because it is an infinite linear combination of indicator functions. – Junyong Kim Dec 9 '17 at 0:02

I think SpaceIsDarkGreen's answer is very good, but I'd like to reformulate one of his examples with a notation that physicists and engineers may relate to more easily. In any case, it made it much easier for me to write it this way so I hope it's useful to others. I'll also argue that the Lebesgue point of view can have a practical use even for integrals of continuous functions.

Consider the (Riemann) integral between $$-1$$ and $$+1$$ of the function $$f(x) = x^2$$:

$$I = \int_{-1}^{1}dx\; f(x) = \int_{-1}^{1}dx\; x^2 = \left.\frac{x^3}{3}\right|_{-1}^{1} = \frac{2}{3}$$

The Lebesgue integral calculates the same things with a very special change of integration variables. I'll write the Lebesgue integral this way:

$$I = \int_{0}^{1} f\;d\mu(f)$$

Basically we're doing what SpaceIsDarkGreen said. For each value of the function $$f \in \left[0,1\right]$$, we find the size $$d\mu(f)$$ of the integration domain that maps to the image $$\left[f,f+df\right]$$. Then we multiply $$f$$ by $$d\mu(f)$$ add up horizontal slices under the curve. To give a practical meaning to our Lebesgue integral, we now need to calculate the size $$d\mu(f)$$. For the integration region above, there are two points on the $$x$$ axis that map to the same value $$f$$, one at $$+x$$ and one at $$-x$$. For one of these points, the size of the domain that maps to $$\left[f,f+df\right]$$ is $$\sqrt{f+df} - \sqrt{f}$$. The function is symmetric so the total size of the domain mapping to $$\left[f,f+df\right]$$ is twice that:

$$d\mu(f) = 2\times\left[\sqrt{f+df}-\sqrt{f}\right] \approx \frac{df}{\sqrt{f}}$$

Here, I used the fact that $$df$$ is small. Putting this back into the Lebesgue integral above, we find that

$$I = \int_{0}^{1} f\;\frac{df}{\sqrt{f}} = \int_{0}^{1} df\;\sqrt{f} = \left.\frac{2}{3}f^{3/2}\right|_{0}^{1} = \frac{2}{3}$$

Of course, we get the same answer. But note that even though we didn't need the Lebesgue integral, we have a very practical result. We proved the following equality:

$$I = \int_{-1}^{1}dx\; x^2 = \int_{0}^{1} dx\;\sqrt{x}$$

and we proved that this equality is not accidental. Here it's a bit trivial, but it may sometimes be the case that even for nice continuous functions, the Lebesgue point of view could give integrals that are easier to solve.

Edit: I still want to emphasize the point made earlier though: this is not the main reason why Lebesgue integration is interesting. But I thought it was worth pointing out that it can give a new point of view on practical integration.