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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?

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    I don't think there is a difference. – 2017-01-03
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    If $f$ is continuous on $[a, b]$ then both integrals are equal. – 2017-01-03
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    (1) The symbol $\int_{D} f(x) \, dx$ encompasses the case where the domain $D$ of integration is not just an interval but any measurable set. So it provides certain convenience in notation. (2) When we are dealing with the Lebesgue measure (or any measure without atoms), then they are equal. On the other hand, if you are dealing with more general measures, then the right-hand side is ill-defined and we should explicitly specify over which set we are integrating. – 2017-01-03
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    "it always assumes $f$ continuous": no, this is not required. – 2017-01-03
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    When you give the definition of definite integral (Theorem of Torricelli-Barrow) I have always taught me that $f$ is continuous. In Mathematical Analysis 1, we never talked about Lebesgue measure. Then they define integrals generally continuous functions and improper integrals. And here we stopped. I remember it very well. – 2017-01-03
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    @Sebastiano Function $f$ can be Riemann integrable on $[a,b]$ without being continuous on $[a,b]$. Actually, the [necessary and sufficient condition for Riemann integrability](https://en.wikipedia.org/wiki/Riemann_integral#Integrability) is that a set of points at which it is not continuous has Lebesgue measure zero – 2017-01-03
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    @ZoranLoncarevic Meanwhile, I thank you all very much. With just heart this condition do not remember absolutely. I might have a theoretical justification (a demonstration, a theoretical explanation, etc ...) with a your answer? – 2017-01-03
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    Each one can add an exhaustive answer in complete form to comments? – 2017-01-03

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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.

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    I have not understood because, $$\int_{[2,3]}\chi_{\mathbb{N}}\; d\tau = 2 $$ and $$ 0 = \int_{(2,3)}\chi_{\mathbb{N}}\; d\tau,$$ Can I have another exhaustive explanations? After why $$\lambda([a,b]) = \lambda((a,b)) = \lambda([a,b)) = \lambda((a,b])\,?$$ If $\varepsilon>0,$ for example $\,\,\lambda([a,b))= \lambda([a,b-\varepsilon])= b-\varepsilon-a$. – 2017-01-03
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    @Sebastiano I've tried to make the counting measure $\tau$ more clear. As for the last thing you mention, $[a,b)$ is not equal to $[a,b-\epsilon]$, but rather to $\displaystyle \lim_{\epsilon \to 0}\;\; [a,b-\epsilon]$. Hence you get $\lim_{\epsilon \to 0} b-\epsilon-a = b-a$. – 2017-01-03
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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.