Riemann-Stieltjes Integral

[jdcasey9]

Homework Statement

If f,g $$\in$$ R_$$\alpha$$[a,b] with f$$\leq$$g, show that $$\int$$_a^bfd$$\alpha$$ $$\leq$$ $$\int$$_a^bgd$$\alpha$$

Homework Equations

Let $$\alpha$$: [a,b] $$\rightarrow$$ R(Real numbers) be increasing. A bounded function f: [a,b] $$\rightarrow$$ R(real numbers) is in R_$$\alpha$$[a,b] if and only if, given $$\epsilon$$$$\succ$$ 0, there exists a partition P of [a,b] such that U(f,P) - L(f,P) $$\prec\epsilon$$.

m=min{m₁,..., m_n}
M=max...
m_i=inf{f(x) : x_i-1$$\leq$$x$$\leq$$x_i}
M_i=sup{f(x) : x_i-1$$\leq$$x$$\leq$$x_i}

The Attempt at a Solution

This problem doesn't seem to be difficult, but I'm running into problems trying to compare f and g:

For f and g: m$$\leq$$ m_i $$\leq$$ L(f,P) $$\leq$$$$\int$$^b_{_a} $$\leq$$$$\int$$^_b_a $$\leq$$ U(f,P) $$\leq$$ M_i $$\leq$$ M

So we can say such things as:

m^f $$\leq$$ m^g for each piece, but can we say that:

m^f $$\leq$$ m^g $$\leq$$ m^f_i $$\leq$$ m^g_i ... etc.

If we can then this is easy to prove. If not, then how do we relate different parts of f and g?
Thanks.

 

[mighty2000]

Thank you for your post. Your approach to the problem is on the right track. To compare f and g, we can use the fact that f\leq g, which means that for every x\in [a,b], f(x)\leq g(x). This implies that for every partition P of [a,b], we have m_i(f)\leq m_i(g) and M_i(f)\leq M_i(g) for all i=1,2,...,n. Therefore, we can write:

L(f,P)=\sum_{i=1}^nm_i(f)\Delta x_i\leq\sum_{i=1}^nm_i(g)\Delta x_i=L(g,P)

and

U(f,P)=\sum_{i=1}^nM_i(f)\Delta x_i\leq\sum_{i=1}^nM_i(g)\Delta x_i=U(g,P)

Since L(f,P)\leq L(g,P) and U(f,P)\leq U(g,P), we can conclude that \int_a^bf(x)d\alpha\leq\int_a^bg(x)d\alpha, as desired.

Hope this helps!


, Scientist