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In my measure theory course, I was taught the following definition of the Lebesgue measure: Let $D_n = \{ x = (x_1, \cdots, x_n) | 2^n x \in \mathbb{Z}^n\}.$ Define the sequence of linear functionals

$$ L_n(f) = \frac{1}{2^{nd}} \sum_{x \in D_n} f(x). $$ Which map compactly supported functions to $\mathbb{C}$.

Then it is possible to prove that the limit of these functionals as $n$ gets large is some other linear functional $L$. By the Riesz representation theorem, this functional $L$ can be expressed as the integral (over $f$'s compact support) with respect to some measure. We call this measure the Lebesgue measure. The notation we used for this measure in $d$ dimensions was $\lambda_d$.

Now my question is the following: how can we make sense of an integral over an unbounded set, with respect to the Lebesgue measure? It seems like this definition is assuming that the functions we're concerned about integrating are compactly supported. How would you integrate a function defined over the real line with respect to the measure $\lambda$? Does this definition need some kind of modification for this more general case? Thank you very much for any answers.

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  • $\begingroup$ You want to assume your $f$ on which the functionals are defined are continuous. The integral is the limit of these functionals for continuous functions of compact support. Once you have your measure, you can try to integrate measurable functions that are not continuous or do not have compact support. But this is not done using the functionals $L_n$ on those functions. $\endgroup$ May 10, 2021 at 21:50
  • $\begingroup$ Then how is it done? That's my question. $\endgroup$ May 10, 2021 at 22:06

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The limit of your sequence defines the Riemann integral of a continuous function. In the theory of Lebesgue integration functions that differ on a null set have equal integrals. With the standard measure on $\mathbb R^n$, the sets $D_n$ are null (as they are countable), and as the countable union of countable sets is countable, $$ \bigcup_{n \in \mathbb N} D_n $$ is also null, so this does not define the Lebesgue integral in the traditional sense, even for compactly supported functions. The standard way to define the Lebesgue measure is to define an outer measure using the volume of cubes (see here) and then restrict to measurable sets to define a measure space. Integrals can then be defined using the monotone convergence theorem and approximations to measurable functions.

In answer to the question, we can, however, start by defining integrals for $f \in C_c(\mathbb R^n)$, the continuous functions with compact support as you describe. We want to define the integrals of non continuous functions and those with non-compact support using some notion of a limit. To do this we can define a norm on $C_c(\mathbb R^n)$ by $$ \|f\| = \lim_{n \to \infty}L_n(|f|) = \int |f| $$ This is not a Banach space as it is not complete (Cauchy sequences don't necessarily have limits). To get around this, we can consider instead the completion of $C_c(\mathbb R^n)$ with respect to the norm $\|\|$, in much the same way as $\mathbb R$ is the completion of $\mathbb Q$, by adding in all the limits. As an aside, from the theory of Lebesgue integration this space is $L^1$, as it can be shown that $C_c(\mathbb R^n)$ is dense in $L^1$.

To define the integral of a non continuous function with non-compact support, $f$, we would view this function as the limit of a Cauchy sequence $f_n \in C_c(\mathbb R^n)$, and it's integral is $$ \int f = \lim_{n \to \infty} \int f_n. $$ provided it is in the space (is the limit of a sequence). More concretely, if $f_n \to f$, consider $$ f_n^m = f_n\chi_{\{|x| \in [0, m]\}} $$ where $\chi$ is the indicator function. Let $f^m = \lim_{n \to \infty}f_n^m = f \chi_{\{|x| \in [0 m]\}}$. Then $f^m$ is a Cauchy sequence in the completion of $C_c(\mathbb R^n)$. Since $\int \cdot = L(\cdot) = \lim_{n \to \infty}L_n(\cdot)$ is bounded linear on our space, $$ \int f = L(f) = \lim_{m\to \infty}L(f^m) = \sum_{m = 1}^\infty L(f\chi_{\{|x| \in [m-1, m]\}}) $$ So, as you would hope, the integral for functions with bounded integrals is just the sum of the integrals of it's compactly supported restrictions.

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  • $\begingroup$ What happens if we have equal parts positive and negative contributions to that last sum you wrote? Eg something like $\int_{\mathbb{R}} x d\lambda$? Would that integral be zero, or do we simply say it doesn't converge? $\endgroup$ May 11, 2021 at 0:36
  • $\begingroup$ The function $f: x \mapsto x$ isn't in the completion of $C_c(\mathbb R)$ (note that the functions $f^m$ in this case would not form a Cauchy sequence). The reason this integral is not $0$ is that by choosing different sets for the characteristic functions we could get the sum in the last equation to converge to any value (or indeed diverge). $\endgroup$
    – Holmes
    May 11, 2021 at 7:49

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