How to compute Riemann-Stieltjes / Lebesgue(-Stieltjes) integral?
Solution 1:
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:
- $F'$ is Riemann-integrable on $[a,b]$, and
- $\displaystyle R\int_a^x F'(t)\,dt = F(x) - F(a)$ 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
- $F$ is continuous at any point where $\phi$ is continuous; and
- $F$ 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
The Riemann-Stieltjes integral of $f_k$ with respect to $\phi$ exists for all $k$; and
The Riemann-Stieltjes integral of $f$ with respect to $\phi$ exists; and
$\displaystyle \text{R-S}\int_a^b f(x)d\phi(x) = \lim_{k\to\infty} \text{R-S}\int_a^b f_k(x)d\phi(x)$.
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
- $F$ is bounded, monotone increasing, absolutely continuous, and differentiable almost everywhere with $F' = f$ almost everywhere;
- 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.
- $\displaystyle \text{L-S}\int_{\mathbb{R}} gd\mu_f = L\int_{\mathbb{R}}gfd\mu = L\int_{\mathbb{R}} gF'd\mu$.
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.
Solution 2:
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.
Solution 3:
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.