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ANALYSIS II
Riemann-Stieltjes Integration: Properties

Theorem. Suppose that a₁, a₂ are non-decreasing, and that f, g are Riemann-Stieltjes integrable with respect to both a₁ and a₂. If c is a nonnegative real number, then

1. ò_a^b c f  da  =  c ò_a^b f da  =  ò_a^b f d(c a)
2. ò_a^b f+g  da  =  ò_a^b f da+ ò_a^b g da
3. ò_a^b f  d(a+b)  =  ò_a^b f da+ ò_a^b f db
4. g £ f  implies  ò_a^b g da  £  ò_a^b f da.
5.  |ò_a^b f da|  £  ò_a^b |f| da  £  ||f||_¥ (a(b) - a(a))
6. ò_a^b f  da = f(s) if f is continuous at s between a and b, and a(x) = I(x-s) where I(x) = 1 for positive x and vanishes otherwise.
7. ò_a^b f  da = å_{n = 1}^N c_n f(s_n)    if  a(x) = å_{n = 1}^N c_n I(x-s_n), f is continuous at each of the s_n's (all of which are assumed to lie in the interval (a,b)), and the c_n's are nonnegative.

Proofs:

Property 1:   We observe that c ³ 0 implies  c a is nondecreasing, M_i( c f ) = c M_i(f) and m_i( c f ) = c m_i(f). Hence U(P;cf,a) = cU(P;f,a) = U(P;f,ca). A similar statement holds for lower sums.

Property 2:   We notice that M_i(f+g) £ M_i(f) + M_i(g) and m_i(f) + m_i(g) £ m_i(f+g) . Hence,

Let e > 0, then since f and g are Riemann-Stieltjes integrable, there exist partitions P₁,P₂ such that

If we let P be a common refinement of P₁ and P₂, then by combining inequalities (1) and (2), we see that

Property 3:   Set g = a+ b and use condition (*) together with the fact that Dg_i = Da_i + Db_i.

Property 4:   This follows directly from the definition of the upper and lower integrals using, for example, the inequality M_i(g) £ M_i(f).

Property 5:   This is proved by applying property 4.) to the inequality

from which it follows that

-ò_a^b |f| da £ ò_a^b f da £ ò_a^b |f| da.

Properties 6-7:   Use the properties above together with our earlier example.

  ^[¯]

Theorem.  If f is Riemann-Stieltjes integrable with respect to a on [a,b], then it is Riemann integrable on each subinterval [c,d] Í [a,b]. Moreover, if c Î [a,b], then

Proof. We set a₁(x) = a(x) on the interval [a,c] and equal to the constant value a(c) on [c,b]. Similarly, we set a₂(x) = a(x)-a(c) if c £ x £ b and define it to vanish on [a,c]. Then the a_j are non-decreasing and a = a₁ + a₂. Note that ò_a^b f da₁ = ò_a^c f da and ò_a^b f da₂ = ò_c^b f da.   ^[¯]