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The definitions do not seem easy to me for computation. For example, Lebesgue(-Stieltjes) integral is a measure theory concept, involving construction for from step function, simple functions, nonnegative function till general function.

I was wondering, in practice, what common ways for computing Lebesgue(-Stieltjes) integral are?

  1. Is it most desirable, when possible, to convert Lebesgue(-Stieltjes) integral to Riemann(-Stieltjes) integral, and Riemann-Stieltjes) integral to Riemann integral, and then apply the methods learned from calculus to compute the equivalent Riemann integral?
  2. What about the cases when the equivalence/conversion is not possible? Is definition the only way to compute Riemann-Stieltjes or Lebesgue(-Stieltjes) integrals?

My questions come from a previous reply by Gortaur

Usually only Lebesgue (Lebesgue-Stieltjes) integrals are used in the probability theory. On the other hand to calculate them you can use an equivalence of Lebesgue-Stieltjes and Riemann-Stieltjes integrals (provided necessary conditions).

Thanks and regards!

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2  
@Tim: You may want to check out Frank Burk's "A Garden of Integrals". It describes the different types and how they relate to one another. In particular, the Lebesgue Integral has an "FTC", and the Lebesgue-Stieltjes Integral can, in many instances, be evaluated much like the Riemann Stieltjes integral can (for example, when $g$ is differentiable). –  Arturo Magidin Apr 11 '11 at 3:38
    
@Arturo: Thanks! I will. By "FTC", you mean Fundamental Theorems of Calculus? Is this en.wikipedia.org/wiki/Differentiation_of_integrals the "FTC" for Lebesgue Integral? Or/and something else? The link does not look like to make the computation easier. –  Tim Apr 11 '11 at 3:41
    
@Tim: I mean the first part of the FTC, which for Riemann Integrals says that if $F'(x) = f(x)$ on $[a,b]$, $f(x)$ continuous, then $\int_a^b f(x)\,dx = F(b)-F(a)$. There are similar theorems for the Lebesgue Integral. The book's in my office, but I'll try to post the theorems tomorrow. –  Arturo Magidin Apr 11 '11 at 3:50
    
You can view most of the book from google. The only condition is that $F$ is absolutely continuous on $[a,b]$. –  GWu Apr 11 '11 at 3:58
    
@Arturo: Thanks! (1) That's nice. I will have the access to the book too, and perhaps just knowing the the numbering of the theorems will be fine and you don't need to type much. (2) Besides FTC, are there other usual ways to calculate the integrals? Are these all possible, integration by parts, by substitution, ... –  Tim Apr 11 '11 at 3:58

3 Answers 3

up vote 15 down vote accepted

Even with the Riemann Integral, we do not usually use the definition (as a limit of Riemann sums, or by verifying that the limit of the upper sums and the lower sums both exist and are equal) to compute integrals. Instead, we use the Fundamental Theorem of Calculus, or theorems about convergence. The following are taken from Frank E. Burk's A Garden of Integrals, which I recommend. One can use these theorems to compute integrals without having to go down all the way to the definition (when they are applicable).

Theorem (Theorem 3.8.1 in AGoI; Convergence for Riemann Integrable Functions) If $\{f_k\}$ is a sequence of Riemann integrable functions converging uniformly to the function $f$ on $[a,b]$, then $f$ is Riemann integrable on $[a,b]$ and $$R\int_a^b f(x)\,dx = \lim_{k\to\infty}R\int_a^b f_k(x)\,dx$$

(where "$R\int_a^b f(x)\,dx$" means "the Riemann integral of $f(x)$").

Theorem (Theorem 3.7.1 in AGoI; Fundamental Theorem of Calculus for the Riemann Integral) If $F$ is a differentiable function on $[a,b]$, and $F'$ is bounded and continuous almost everywhere on $[a,b]$, then:

  1. $F'$ is Riemann-integrable on $[a,b]$, and
  2. $\displaystyle R\int_a^x F'(t)\,dt = F(x) - F(a)$ for each $x\in [a,b]$.

Likewise, for Riemann-Stieltjes, we don't usually go by the definition; instead we try, as far as possible, to use theorems that tell us how to evaluate them. For example:

Theorem (Theorem 4.3.1 in AGoI) Suppose $f$ is continuous and $\phi$ is differentiable, with $\phi'$ being Riemann integrable on $[a,b]$. Then the Riemann-Stieltjes integral of $f$ with respect to $\phi$ exists, and $$\text{R-S}\int_a^b f(x)d\phi(x) = R\int_a^b f(x)\phi'(x)\,dx$$ where $\text{R-S}\int_a^bf(x)d\phi(x)$ is the Riemann-Stieltjes integral of $f$ with respect to $d\phi(x)$.

Theorem (Theorem 4.3.2 in AGoI) Suppose $f$ and $\phi$ are bounded functions with no common discontinuities on the interval $[a,b]$, and that the Riemann-Stieltjes integral of $f$ with respect to $\phi$ exists. Then the Riemann-Stieltjes integral of $\phi$ with respect to $f$ exists, and $$\text{R-S}\int_a^b \phi(x)df(x) = f(b)\phi(b) - f(a)\phi(a) - \text{R-S}\int_a^bf(x)d\phi(x).$$

Theorem. (Theorem 4.4.1 in AGoI; FTC for Riemann-Stieltjes Integrals) If $f$ is continuous on $[a,b]$ and $\phi$ is monotone increasing on $[a,b]$, then $$\displaystyle \text{R-S}\int_a^b f(x)d\phi(x)$$ exists. Defining a function $F$ on $[a,b]$ by $$F(x) =\text{R-S}\int_a^x f(t)d\phi(t),$$ then

  1. $F$ is continuous at any point where $\phi$ is continuous; and
  2. $F$ is differentiable at each point where $\phi$ is differentiable (almost everywhere), and at such points $F'=f\phi'$.

Theorem. (Theorem 4.6.1 in AGoI; Convergence Theorem for the Riemann-Stieltjes integral.) Suppose $\{f_k\}$ is a sequence of continuous functions converging uniformly to $f$ on $[a,b]$ and that $\phi$ is monotone increasing on $[a,b]$. Then

  1. The Riemann-Stieltjes integral of $f_k$ with respect to $\phi$ exists for all $k$; and

  2. The Riemann-Stieltjes integral of $f$ with respect to $\phi$ exists; and

  3. $\displaystyle \text{R-S}\int_a^b f(x)d\phi(x) = \lim_{k\to\infty} \text{R-S}\int_a^b f_k(x)d\phi(x)$.

One reason why one often restricts the Riemann-Stieltjes integral to $\phi$ of bounded variation is that every function of bounded variation is the difference of two monotone increasing functions, so we can apply theorems like the above when $\phi$ is of bounded variation.


For the Lebesgue integral, there are a lot of "convergence" theorems: theorems that relate the integral of a limit of functions with the limit of the integrals; these are very useful to compute integrals. Among them:

Theorem (Theorem 6.3.2 in AGoI) If $\{f_k\}$ is a monotone increasing sequence of nonnegative measurable functions converging pointwise to the function $f$ on $[a,b]$, then the Lebesgue integral of $f$ exists and $$L\int_a^b fd\mu = \lim_{k\to\infty} L\int_a^b f_kd\mu.$$

Theorem (Lebesgue's Dominated Convergence Theorem; Theorem 6.3.3 in AGoI) Suppose $\{f_k\}$ is a sequence of Lebesgue integrable functions ($f_k$ measurable and $L\int_a^b|f_k|d\mu\lt\infty$ for all $k$) converging pointwise almost everywhere to $f$ on $[a,b]$. Let $g$ be a Lebesgue integrable function such that $|f_k|\leq g$ on $[a,b]$ for all $k$. Then $f$ is Lebesgue integrable on $[a,b]$ and $$L\int_a^b fd\mu = \lim_{k\to\infty} L\int_a^b f_kd\mu.$$

Theorem (Theorem 6.4.2 in AGoI) If $F$ is a differentiable function, and the derivative $F'$ is bounded on the interval $[a,b]$, then $F'$ is Lebesgue integrable on $[a,b]$ and $$L\int_a^x F'd\mu = F(x) - F(a)$$ for all $x$ in $[a,b]$.

Theorem (Theorem 6.4.3 in AGoI) If $F$ is absolutely continuous on $[a,b]$, then $F'$ is Lebesgue integrable and $$L\int_a^x F'd\mu = F(x) - F(a),\qquad\text{for }x\text{ in }[a,b].$$

Theorem (Theorem 6.4.4 in AGoI) If $f$ is continuous and $\phi$ is absolutely continuous on an interval $[a,b]$, then the Riemann-Stieltjes integral of $f$ with respect to $\phi$ is the Lebesgue integral of $f\phi'$ on $[a,b]$: $$\text{R-S}\int_a^b f(x)d\phi(x) = L\int_a^b f\phi'd\mu.$$


For Lebesgue-Stieltjes Integrals, you also have an FTC:

Theorem. (Theorem 7.7.1 in AGoI; FTC for Lebesgue-Stieltjes Integrals) If $g$ is a Lebesgue measurable function on $R$, $f$ is a nonnegative Lebesgue integrable function on $\mathbb{R}$, and $F(x) = L\int_{-\infty}^xd\mu$, then

  1. $F$ is bounded, monotone increasing, absolutely continuous, and differentiable almost everywhere with $F' = f$ almost everywhere;
  2. There is a Lebesgue-Stieltjes measure $\mu_f$ so that, for any Lebesgue measurable set $E$, $\mu_f(E) = L\int_E fd\mu$, and $\mu_f$ is absolutely continuous with respect to Lebesgue measure.
  3. $\displaystyle \text{L-S}\int_{\mathbb{R}} gd\mu_f = L\int_{\mathbb{R}}gfd\mu = L\int_{\mathbb{R}} gF'd\mu$.


The Henstock-Kurzweil integral likewise has monotone convergence theorems (if $\{f_k\}$ is a monotone sequence of H-K integrable functions that converge pointwise to $f$, then $f$ is H-K integrable if and only if the integrals of the $f_k$ are bounded, and in that case the integral of the limit equals the limit of the integrals); a dominated convergence theorem (very similar to Lebesgue's dominated convergence); an FTC that says that if $F$ is differentiable on $[a,b]$, then $F'$ is H-K integrable and $$\text{H-K}\int_a^x F'(t)dt = F(x) - F(a);$$ (this holds if $F$ is continuous on $[a,b]$ and has at most countably many exceptional points on $[a,b]$ as well); and a "2nd FTC" theorem.

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Thanks! That is really helpful! –  Tim Apr 11 '11 at 21:42
1  
@Tim: You're welcome. The point is that, just as with limits, and derivatives, and Riemann integrals, we seldom want to go back all the way to the definition in order to compute integrals; instead, we want to settle some basic instances, and then apply a bunch of theorems that tell us how to compute the results for more complicated functions built up from simpler ones (when possible). There may (and are) some instances when one does need to go all the way back to the definition, but we usually don't. –  Arturo Magidin Apr 11 '11 at 21:45
    
Thanks! Generally, when talking about Lebesgue integral (in narrow sense), Lebesgue-Stieltjes integral or Riemann-stieltijes integral, are the integrands all real-valued Borel measurable functions defined on $\mathbb{R}$, not on $\mathbb{R}^n$, nor on other more general measure space? I know in Rudin's real and complex analysis, Lebesgue integral has a wider sense and applies to real-valued integrands defined on general measure space. –  Tim Apr 12 '11 at 1:04
    
@Tim: The "standard" are functions that are Lebesgue measurable (more general than Borel measurable). We can do it with values on any measure space, though $\mathbb{R}$ and $\mathbb{R}^n$ (with the Lebesgue product measure on the latter) are the most common. I took a course in which we did the entire development of Lebesgue integration with Banach-space-valued functions, and there was absolutely no difficulty in generalizing from $\mathbb{R}$ to that situation. –  Arturo Magidin Apr 12 '11 at 1:55

The answer depends on what do you mean by practice. In the most common situations the integral of a function $f$ can be usually found by comparing the $f$ to the functions $g$ for which we already know what integral looks like. For the large class of commonly used functions all the notions of the integrals coincide, so it is not necessary to convert from one integral to another, when you have already calculated one variant.

If for the certain function only one notion of integral applies, the comparison principle still applies. The classical example is the Dirichlet function for which Riemman integral does not exist. On the other hand this is zero almost everywhere, and Lebesgues integral is the same for functions which differ on the set of Lebesgue measure zero. So we get the Lebesgue integral of Dirichlet function is the Lebesgue integral of zero. For zero function Rieman and Lebesgue integrals coincide so we can calculate which ever is easier.

Another trick is to construct the sequence of functions with known integrals which converge to the function of interest. Then integral is usually the limit of the corresponding integrals. This of course does not apply for all types of convergence and functions.

To sum up, the definition is not the only way to calculate the integral. The definition is used to calculate the integrals of the most simple functions, the rest is usually calculated by manipulating the function or integral to already known solution.

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It will be better if you will provide the area (or the problem) which leads you to the calculation of this integrals. From the computational point of view there are two "types" of integrals which lead you to correspondent two general methods of its computation. They are dependent on the distribution $Q$. Let us consider the case of $\mathbb{R}$.

The first type is an integral of an absolutely continuous distributions $Q$ - i.e. of such that $Q(dx) = h(x)\,dx$ for function $h$ which is a density function. These integrals often are calculated like a Riemann integrals (using correspondent methods).

All other 1-dimensional integrals for the computations can be reduced to the previous case. For the cumulative distribution function $g$ (which always exists) you can write $$ dg(x) = h(x)dx + \sum\limits_{i=1}^n p_i\delta(x-x_i) $$ where $\delta(x)$ is a Dirac function.

Then for the continuous function $f$ $$ \int\limits_{\mathbb{R}}f(x)\,dg(x) = \int\limits_{-\infty}^\infty f(x)h(x)\,dx +\sum\limits_{i=1}^n p_if(x_i). $$

This also will be the case if $f$ has countably many discontinuities which do not coincide with the sequence $(x_i)$ of massive points.

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Thanks! The area can be probability theory, but can be in complex analysis and manifold. I was wondering what types of integral are used in complex analysis, such as for Fourier transformation? Is integral over manifold called integral of differential forms? Are the integrals from these two areas special cases of Lebesgue integral over a general measure space? –  Tim Apr 11 '11 at 14:01
    
I think that integration of a differential from deals more with a Riemann integral than with the Lebesgue one. The idea is the following - the Riemann construction uses partition of the domain of the function (which then easily extended on the limit case with non-compact sets). On the other hand, to define the Lebesgue integral one usually deals with the partition of the codomain and measures of pre-images. The Dirichlet function is a good example where partition of the codomain is easy, on the other hand, the Riemann integral is undefined. –  Ilya Apr 11 '11 at 15:54

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