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?

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

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 $$ \int_{\mathbb R^2} f(w)\,dw. $$


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.


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