Extended integral

Question

What are the differences between (supposed true all hypotheses),

$$\int_{[a,b]}f(x)\text{d}x \quad\text{and} \quad\int_{a}^{b}f(x)\text{d}x\,\,?$$

This is an old topic of mathematical analysis of 23 years ago when I was a university student. When you give the definition of integral it always assumes $f$ continuous. What sense does it make to define the symbol of the extended integral

$$\int_{[a,b]}f(x)\text{d}x$$ if you do not need it?

Answer 1

The reason $$\int_{[a,b]} f(x)dx$$ is defined, is because integration is not limited to the "standard" Riemann or Lebesgue integration! Integrating over the reals is usually done using something called the Lebesgue measure $\lambda$, which (in very simple terms) measures the length of a set.

In particular, for an interval between $a$ and $b$, the Lebesgue measure is defined as $$\lambda([a,b]) = \lambda((a,b)) = \lambda([a,b)) = \lambda((a,b]) = b-a.$$

Since for this measure it doesn't matter if $a$ and $b$ are inside or outside the interval, we can write (without ambiguity) $$\int_a^b f(x)dx.$$

In general, we define integration with respect to a measureable set $D$ and a measure $\mu$: $$\int_D f \; d\mu$$ Without going into much detail what all of this is, let me define another measure, the counting measure: $$\tau: D \mapsto \left\{\begin{array}{ll} |D| & \text{ if $D$ is a finite set}\\ \infty & \text{ if $D$ is an infinite set} \end{array}\right.$$ Using this measure to integrate, an integral of a positive function $f$ can be evaluated as follows: $$\int_D f \;d\tau = \left\{\begin{array}{ll}\sum\limits_{x \in D}f(x) & \text{ if $f$ has countable support}\\ +\infty & \text{ if $f$ has uncountable support } \end{array}\right.$$ For example, take the indicator function $$\chi_\mathbb{N}: \mathbb{R} \to \{0,1\}: x \mapsto \left\{\begin{array}{ll}1 & \text{ if } x \in \mathbb{N}\\ 0 & \text{ if } x \notin \mathbb{N}\end{array}\right.$$ Then integrating over an interval can depend on the boundaries of said interval: $$\begin{align}\int_{[2,3]}\chi_{\mathbb{N}}\; d\tau &= \sum\limits_{x \in [2,3]}\chi_{\mathbb{N}}(x) = 1 + 0 + 0 + \cdots + 0 + 1 = 2\\ \int_{(2,3)}\chi_{\mathbb{N}}\; d\tau &= \sum\limits_{x \in (2,3)}\chi_{\mathbb{N}}(x) = 0 + 0 + \cdots + 0 = 0\\ \int_{(2,3]}\chi_{\mathbb{N}}\; d\tau &= \sum\limits_{x \in (2,3]}\chi_{\mathbb{N}}(x) = 0 + 0 + \cdots + 0 +1= 1\\ \int_{[2,3)}\chi_{\mathbb{N}}\; d\tau &= \sum\limits_{x \in [2,3)}\chi_{\mathbb{N}}(x) = 1 + 0 + 0 + \cdots + 0 = 1\\ \end{align}$$ So in this case we need this notation, because the notation $$\int_2^3 \chi_\mathbb{N} \;d\tau$$ would be ambiguous.

Answer 2

If we are talking about Riemann integral here, for an arbitrary subset $A \subset [a,b]$ $$\int_A f dx = \int_a^b f \chi_A dx$$ where $\chi_A : [a,b] \to \{0,1\}$ is characteristic function of set $A$ given by $$\chi_A(x) = \begin{cases} 1, & x \in A \\ 0, & x \notin A \end{cases}$$ provided that function $f \chi_A$ is Riemann integrable on $[a,b]$. With that in mind, for every function $f$ that is Riemann integrable on $[a,b]$ $$\int_{[a,b]} f dx = \int_{(a,b]} f dx = \int_{[a,b)} f dx = \int_{(a,b)} f dx = \int_a^b f dx$$ since value of Riemann integral does not depend on the value of function in any finite number of points.

So, as far as Mathematical Analysis 1 syllabus is concerned, the answer is that there is no diference. In general, $$\int_{[a,b]} f dx$$ might also mean the Lebesgue integral of function $f$ over interval $[a,b]$ which for Riemann integrable function $f$ coincide with Riemann integral, but is also defined for some functions that are not Riemann integrable.