Consider: $\int \int_A 1 \ dx dy, \quad$ where $A = [3,3]\times\mathbb{R}^2$.

I know this is probably an extremely basic question, but I've forgotten everything from my multivariable calculus class many years ago, and I'm having trouble with some conceptual questions regarding the above problem.

First question: How does one evaluate this integral? Intuitively, I feel like it should be equal to 0 since you're basically taking the area of a line. And indeed:

$\int^\infty_{-\infty} \int^3_3 1 \ dxdy = \int^\infty_{-\infty} 0 dy = 0$.

But when I switched the order of integration, I obtained:

$\int^3_3 \int^\infty_{-\infty} 1 \ dy dx, \quad$ which does not converge.

Here I thought that this can't be right, and I know that in the one variable case, an infinite integral can be expressed as a limit. So I wrote the above again as:

$\lim_{a \rightarrow \infty} \lim_{b \rightarrow -\infty}\int^a_{b} \int^3_3 1 \ dxdy = \lim_{a \rightarrow \infty}\lim_{b \rightarrow -\infty} \int^3_3 \int^a_{b} 1 \ dydx = 0$.

But I'm not entirely convinced whether I can use this one-variable concept of limits in the multi-variable case, and furthermore that I can pull this limit out of the integral. What is the best and most rigorous way to evaluate this (or show that it does not exist)?

Second question: In a more general sense, I understand Fubini's theorem to imply that if

$\int \int_A f(x,y) \ dx dy, \quad$ exists, then assuming that you have the correct bounds for integration,

$\int \int f(x,y) \ dx \ dy = \int \int f(x,y) \ dy \ dx$

If my understanding of Fubini's theorem is true (and if not please correct me), does the converse hold? i.e. if:

$\int \int f(x,y) \ dx \ dy = \int \int f(x,y) \ dy \ dx$

then does

$\int \int_A f(x,y) \ dx dy, \quad$ exists?

I hope those questions made sense, and thank you in advance!

Note: I haven't yet learned Lebesgue integration, nor any measure theory, so if it possible to answer the question using only Riemann integrals, and no measure theory, that would be great!

  • $\begingroup$ $1$ isn't integrable over $\{3\}\times\mathbb{R}$ since if we take the integral over just $\mathbb{R}$ the integral will never converge meaning fubini's theorem doesn't apply. This means that $\int\int_A dxdy$ is not well defined. $\endgroup$
    – J.V.Gaiter
    Dec 3, 2020 at 21:35

1 Answer 1


For your first question: over in Riemann-land, we define integrals over unbounded domains as the limit of integrals over bounded domains(this is not the case when working from a measure-theoretic standpoint). Thus, for any $C,D > 0$, let's consider $A_C^D = [3,3] \times [-C,D]$ (I'm assuming the $\mathbb{R}^2$ is a typo, as you seem to be using $\mathbb{R}$ in its place after the first line).

Then we have $$\int_{A}1 = \lim\limits_{C\to\infty}\lim\limits_{D\to\infty}\int_{A_C^D}1,$$ by definition.

So, let's evaluate the integral inside the limits. Since $1$ is continuous, we need only exhibit one sequence of Riemann sums that we can compute the limit of, and we can be sure that all such sequences converge to the same limit.

So, keeping life easy for ourselves, let's just use the constant sequence $A_n = \{A_C^D\}$. We obviously have $A_C^D = \bigcup_{B \in A_C^D}B$ for all $n$, and $\Delta A_C^D = 0$ for all $n$, so in particular, $\Delta A_n \to 0$ as $n \to \infty$.

Thus, we have $$\int_{A_C^D}1 = \lim\limits_{n\to\infty}1\left(\frac{C+D}{2}\right)\Delta A_C^D = \lim_{n\to\infty}0 = 0,$$ (where $1$ is the constant function and the brackets denote function evaluation) and hence $$\int_A1 = \lim_{C\to \infty}\lim_{D \to \infty}0 = 0.$$


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