3
$\begingroup$

I have a measurable function $f : \mathbb{R}^d \to \mathbb{R}$. Let $E$ be a measurable subset of $\mathbb{R}^d$. Then then $$\int_{E} f(x) \, dx = \int f(x) \chi_E (x) \, dx.$$

If we are taking an integral over $\mathbb{R}^d$, shouldn't we have multiple integrals? Is this just a short hand notation that is never explicitly mentioned?

$\endgroup$
  • $\begingroup$ I don't think so, if your measure is $n$-dimensional Lebesgue measure $\endgroup$ – MPW Oct 10 '14 at 23:49
5
$\begingroup$

The Lebesgue integral of a function $f$ on a measure space $X$ with measure $m$ is denoted $$ \int_X f\, dm\text{ or }\int_X f(x)\,dm(x) \text{ or (especially in probability theory) }\int_X f(x) m(dx). $$ If it is understood which measure is involved, it may be denoted $$ \int_X f(x) \, dx. $$ If the space happens to be $\mathbb R^d$, that is no exception. Perhaps what is "never explicitly mentioned" is that $\mathbb R^d$ not an exception to this way of writing integrals. What is explicitly mentioned would depend on what source you're reading.

Notice that an integral with respect to $d$-dimensional Lebesgue measure is not an integral of an integral with respect to lower-dimensional measures. It is defined by means of measures on $d$-dimensional space, not by measures on lower-dimensional spaces.

However, it is nonetheless called a multiple (or double, or triple, etc., as the case may be) integral, as distinguished from an iterated integral.

One may write $$ \int_{\mathbb R^2} f(x,y)\,d(x,y) \tag 1 $$ for the integral with respect to $2$-dimensional Lebesgue measure, a "double integral", or $$ \int_{\mathbb R} \int_{\mathbb R} f(x,y)\,dx\,dy $$ for the "iterated" (as opposed to "double") integral, which is $$ \int_{\mathbb R} \left( \int_{\mathbb R} f(x,y)\,dx\right)\,dy, $$ an integral of an integral, each with respect to $1$-dimensional Lebesgue measure.

Instead of $(1)$, one might denote the pair $(x,y)$ by a single letter $w=(x,y)$, and write the integral as $$ \int_{\mathbb R^2} f(w)\,dw. $$

$\endgroup$
0
$\begingroup$

It's like $\int_{E_1\times\cdots \times E_n} f(x_1,\ldots,x_n) \, d(x_1)\cdots d(x_n)$ where $E_1\times\cdots \times E_n=E$. Here you can use Tonelli's and Fubini's theorems.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.