How to prove this sequence of Riemann sums -> Inf a^b f(x) dx?

Question

Suppose f(a, b] --> R is continuous, monotonically decreasing, goes to oo as x --> a+ but its improper Riemann integral exists over (a, b]. Like f(x) = 1/sqrt(x) over (0, 1].

Let S_n be a sequence of Riemann sums taken over [a, b], associated with partitions P_n, in such a way that: (1) The norm of P_n (length of the largest interval) --> 0 and (2) For each interval of P_n, the tag point, at which f is evaluated, is its the interval right end point.

Show that S_n --> Int a^b f(x) dx

Are these conditions too rigid? Can any of them be dropped without compromising the conclusion?

Thank you

Answer 1

I think Steiner's answer was super-duper-overkill here; this can all be addressed fairly straightforwardly from a Riemannian approach.

First of all, what is your definition of "its improper Riemann integral exists"? Typically, the definition of "Riemann integrable" is that lim (||Pn||→0) Sn exists, where Sn = ∑||I_n||∙t_n, where I_n is the n'th interval and t_n is an arbitrary point in I_n, and ||Pn|| is the max of ||I_n||.

From this definition, your conclusion is fairly immediate, since the right-hand endpoint is just a special case of t_n in the definition. But, as I said, it depends I guess on how you're defining "the improper integral exists."

I'm guessing you mean something like lim (ε→0)∫fdx on the interval [a+ε,b] exists. But then to get your result, just note that since the limit exists (call it L) and f is monotone decreasing, you know that
f(x)(x-a) ≤ L for ANY x in (a,b]. This means in particular that the first interval in any partition will be bounded, so you're set.
[Note: I'm assuming f is non-negative. If not, just take the function f(x)+f(b); since f is monotone decreasing, this will do the trick.]
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As for your conditions... You certainly don't need continuity; the fact that f is monotone and integrable is good enough. (In fact, monotone → almost-everywhere continuous.) And, obviously, you really only need t_1 (the first "tag point) to be the right-hand endpoint; the rest of the tags don't matter. In fact, I'm not even sure t_1 needs to be the endpoint, because if it isn't you can always give a refinement splitting I_1 into [a,t_1] and [t1,x1].
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Answer 2

Based on the given conditions, I think I can prove this using some measure theory.

We can assume f doesn't take on negative values. Otherwise, split [a, b] into 2 intervals, the first where f is nonnegative, the other where it's negative, and aplly this proof twice, considering -f in the case of the second interval.

For each n, the Riemann sum S_n can be regarded as the Lebesgue integral, with respect to the Lebesgue measure m, of a simple function phi_n (in this case, a step function) whose finite range is composed of the values of f at the tag points.

Since f is decreasing and we're assuming it's non negative, its is easy to see that 0 <= phi_n <= f and that phi_n is an increasing sequence of simple functions. And since, in addition, f is continuous and norm(P_n) --> 0, it follows phi_n --> f.

So, according to the monotone convergence theorem, it follows that

lim Integral phi_n dm = lim S_n = Integral f dm, the integrals taken over (a, b]. (1)

Since the improper Riemann integral of f exists over (a, b], its proper integral exists over [c, b] for every c in (a, b). Therefore, for such values of c, we have

(Lebesgue) Integral f dm =Integral f(x) dx (Riemann) over [c, b]. (2)

According to the properties of the Lebesgue integral, we have, in addition, that

lim (c --> a+) Integral (over [c, b]) f dm = Integral (over (a, b]) f dm (3) and, since the improper Riemann integral of exists over (a, b],

lim ( c--> a+) Integral c^b f(x) dx = Integral a^b f(x) dx (4)

Combining (1), (2), (3) and (4), we finally get

lim S_n = Integral a^b f(x) dx, as desired.

I'm not sure if you can drop some of ths assumptions, I'll think about this. Maybe continuity is not required.

Answer 3

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