# Why is the Lebesgue Integral defined through integrals of simple functions?

I'm reading the Measure Theory and Integration chapter from Terrance Tao's Analysis-2. In the chapter of Lebesgue Integration (11), initially the integral is introduced for positive valued functions and then generalized to negative valued function by first defining a way to split up the negative and positive value of function.

What I don't understand is, why can't we directly integrate functions which take any value in the Lebesgue Integration? Could someone give an intuition to me on why we do this?

I understand stuff like defining the measure function, why we need measurable sets , measurable functions etc but this point I am confused as heck.

Edit : I found this relevant MO post https://mathoverflow.net/q/25161/159957

• If you think you have a better way of defining the integral, you should tell us in some detail what it is. It's hard to say what's wrong (if anything) with your preferred approach if we don't know what it is.
– bof
Commented Jul 30, 2022 at 8:47
• Well, if you want your integral to be linear, then to integrate any complex valued function, it suffices to define the integral on nonnegative functions, and then just extend it to complex valued functions by Linearity. So really the approach is very simple. You can then bootstrap even further by defining the Bochner integral of functions that take values in a separable Banach space. See mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/… Commented Jul 30, 2022 at 22:44
• @bof I do not think I have the level of math genius level like lebesgue. I tried pointing out my issue, the purpose was to get an explanation which peasants like me could understand.
– Babu
Commented Aug 1, 2022 at 5:24
• Edit: This is mentioned in a single line of tao if you read very carefully. I missed it, but anyways insight gained from the post was worth it
– Babu
Commented Apr 6, 2023 at 22:40

You don't have to do it this way. This is just one (convenient) approach. It is convenient because doing it this way allows us to very easily handle $$\infty$$.

There are other textbooks (e.g Lang, Amann and Escher) which do things differently. Fix a measure space $$(X,\mathfrak{M},\mu)$$, and a real/complex finite-dimensional (for simplicity) normed vector space $$V$$. We always equip $$V$$ with the topology induced by the norm, and the Borel $$\sigma$$-algebra associated to this topology. Fun fact: since $$V$$ is finite-dimensional, all norms on $$V$$ generate the same topology, and hence the same Borel $$\sigma$$-algebra.

Approach 1: Directly Integrating $$V$$-valued functions.

Now we can define a $$V$$-valued simple function on $$X$$ in the obvious way, namely a measurable function $$s:X\to V$$ with a finite image set. Then, letting $$v_1,\dots, v_k$$ be the distinct elements of the image, we can decompose it as \begin{align} s=\sum_{i=1}^k\mathbf{1}_{s^{-1}(\{v_i\})}v_i, \end{align} i.e it is exactly what we expect: a finite combination of indicators. Then, we define $$\int_Xs\,d\mu$$ in the obvious way, provided each of the sets defining the indicators has finite measure. Then, we define $$\mathcal{L}^1(X,\mu,V)$$ to be the set of measurable $$f:X\to V$$ for which there is a "$$L^1$$-Cauchy-sequence of simple functions" which approximates; more precisely the requirement is that there is a sequence $$\{s_n\}_{n=1}^{\infty}$$ of $$V$$-valued measurable functions such that

1. we have $$s_n\to f$$ pointwise $$\mu$$-a.e. (approximate $$f$$)
2. For every $$\epsilon>0$$, there is an $$N\in\Bbb{N}$$ such that for all $$n,m\geq N$$, we have $$\int_X\|s_n-s_m\|_V\,d\mu <\epsilon$$. ($$L^1$$-Cauchy).

A note about condition 2: you obviously need to see it's well-defined in the following sense: you need to know that the space of simple functions forms a vector space, so that $$s_n-s_m$$ is again a $$V$$-valued simple function. Also, $$\|s_n-s_m\|_V$$ is a $$[0,\infty)\subset\Bbb{R}$$-valued simple function. THese are the sorts of basic stuff treated more systematically in textbooks which is why I'm glossing over them.

Condition 2 implies that $$\{\int_Xs_n\,d\mu\}_{n=1}^{\infty}$$ is a Cauchy-sequence in $$V$$, so by completeness of $$V$$, it follows that this sequence has a limit, and we define $$\int_Xf\,d\mu:=\lim\limits_{n\to\infty}\int_Xs_n\,d\mu$$. One final well-definition check is required: we need to show that if we take a different sequence $$\{s_n'\}_{n=1}^{\infty}$$ satisfying $$(1)$$ and $$(2)$$, then we get the same limit. This is again proven in textbooks.

So, the long story short is that we have directly defined integration of $$V$$-valued measurable functions, as limits of integrals of an approximating sequence of simple functions. This idea must be extremely familiar to you from Riemann integration as well: the Riemann sums play the role of the simple functions, and the full Riemann integral is defined via a careful limiting procedure.

This approach doesn't rely on the total order of $$[0,\infty]$$, and with some tweaks can even handle infinite-dimensional Banach spaces, so in that sense may be desirable.

Approach 2: A Hybrid.

However, as I said in the beginning, being able to handle $$\infty$$ is a very convenient thing, because it makes the 3 major limit theorems (MCT, Fatou's lemma, and DCT) extremely easy to state. For this reason, it is good to first define integration of $$[0,\infty]$$ valued functions. Then, rather than splitting things up into positive and negative parts, followed by real/imaginary parts, we can proceed as follows:

• Define $$\mathscr{L}^1(X,\mu,V)$$ as the set of measurable $$f:X\to V$$ such that $$\|f\|_1:=\int_X\|f\|_V\,d\mu<\infty$$. Note that $$f:X\to V$$ is measurable and $$\|\cdot\|_V:V\to [0,\infty)$$ is continuous, so the composition $$\|f\|_V:X\to [0,\infty)$$ is measurable, so the above integral makes sense. Note that $$\|\cdot\|_1$$ defines a semi-norm on $$\mathscr{L}^1(X,\mu,V)$$, so it allows us to give a topology on $$\mathscr{L}^1(X,\mu,V)$$.
• Define $$\mathcal{S}(X,\mu,V)$$ to be the space of all simple integrable functions.
• Prove that $$\mathscr{L}^1(X,\mu,V)$$ is a vector space over the same field as $$V$$, and that $$\mathcal{S}(X,\mu,V)$$ is a subspace. In fact, the simple functions $$\mathcal{S}(X,\mu,V)$$ are dense in $$\mathscr{L}^1(X,\mu,V)$$, with respect to the above semi-norm.
• The integrtion map $$\int_X(\cdot)\,d\mu:\mathcal{S}(X,\mu,V)\to V$$ is linear and continuous with respect to the semi-norm on the domain, and the usual norm on $$V$$, i.e $$\|\int_Xs\,d\mu\|\leq \int_X\|s\|_V\,d\mu$$. For simple functions, this is a very trivial consequence of the triangle inequality.
• Finally, we show that the integration map has a unique continuous linear extension $$\int_X(\cdot)\,d\mu:\mathscr{L}^1(X,\mu,V)\to V$$.

I should remark that for a finite-dimensional vector space, this way of stating things is a little unnecessarily abstract (because once we define integrals for $$[0,\infty]$$ functions, by splitting into positive and negative parts, we can define it for all real functions, then by dealing with each component of $$f$$ relative to a basis on $$V$$, we can define integrability of $$f$$), but for infinite-dimensional (separable) Banach spaces, this is necessary.

Anyway, I'm skipping over all proofs here, and glossing over several details because it's too much to write in an MSE post. But regardless of the "implementation details", the basic idea is to first define integrals for simple functions, and then extend by a limiting argument, the integral for other functions.

• Okay, thank you for the answer ans continued support. I have a few more stuff to learn before I can comprehend it but, again, thanks.
– Babu
Commented Aug 1, 2022 at 5:25
• I hope this is not going too off topic, but how would one define the simple functions if $V$ is infinite dimensional (but still separable)? I've been going over some notes that explain the construction but I struggle to see the motivation for how they define the coefficient of each indicator function when defining simple functions to approximate $f$. Commented Jun 5 at 6:48
• @CBBAM the definition is unchanged: for any two measurable spaces $X,Y$, a you can consider a map $s:X\to Y$ which is measurable (preimage of measurable sets are measurable) and whose image is a finite set. Now, suppose $Y=V$ is a Banach space; then you equip it with the Borel sigma-algebra. So, now we know what a simple function $s:X\to V$ means. Next, suppose $(M,\mathfrak{M},\mu)$ is a measure-space (not just measurable space) and $V$ is Banach. Then, we would like to add one extra condition for the purposes of integration, that is $\mu(s^{-1}(V\setminus\{0\}))<\infty$. Commented Jun 5 at 7:04
• This just ensures that if $s=\sum_{i=1}^n\chi_{E_i}v_i$, with each $v_i\neq 0$ then the integral $\int_Xs\,d\mu:=\sum_{i=1}^n\mu(E_i)\cdot v_i$ is actually a well-defined element of $V$. Anyway just read Lang or Amann-Escher; they cover everything from scratch. Commented Jun 5 at 7:05
• @peek-a-boo Thank you! Commented Jun 5 at 15:51

In a way this is because addition of plus infinity and minus infinity is not defined.

The Lebesgue measure assigns nonnegative and possibly infinite values to sets, and addition of such values is well-defined (and hence so is the additivity of the Lebesgue measure). Integration of nonnegative functions similarly is well-defined, because this also is an addition of nonnegative and possibly infinite values (which each are the product of the function value times the Lebesgue measure of a set); e.g., for a simple nonnegative-valued function $$\begin{equation*} f = \sum^n_{i=1}a_i 1_{A_i}, \end{equation*}$$ where the $$a_i$$ are nonnegative numbers and the $$A_i$$ are disjoint measurable sets, the integral $$\begin{equation*} \int f(x)dx = \sum^n_{i=1}a_i\lambda(A_i) \end{equation*}$$ (where $$\lambda$$ is the Lebesgue measure) is well-defined, and this is generalised to arbitrary nonnegative-valued measurable functions through convergence arguments. This definition, however, does not directly generalise to real-valued measurable functions, because it would require a definition of plus infinity and minus infinity, which usually is left undefined; e.g. the function that is minus one up until zero and plus one above zero, i.e., $$\begin{equation*} f = -1_{(-\infty,0]}+1_{(0,\infty)} \end{equation*}$$ has no well-defined Lebesgue integral because $$-\infty + \infty$$ is not defined.