Improper integral convergence proof

[Hoblitz]

Homework Statement

Let $[a, b)$ be an interval in the reals, with $-\infty < a < b \leq \infty$, and let $\alpha: [a,b) \to \mathbb{R}$ be monotone increasing. Suppose that $f: [a,b) \to \mathbb{R}$ is a function such that for each $c \in (a,b)$, $f$ is integrable over $[a,c ]$ with respect to $\alpha$ (in particular f is bounded on every compact subinterval of $[a, b)$, although f may not be bounded on all of $[a, b)$).

Now for the actual problem...

Prove that if $F(x) = \int_a^x f(x) d\alpha$ is a bounded function $[a, b) \to \mathbb{R}$ and $g: [a, b) \to \mathbb{R}$ is non-negative valued, monotone decreasing, integrable with respect to $\alpha$ over each interval $[a, c]$ with a < c < b, and $\lim_{x \to b^{-}} g(x) = 0$, then

$$\int_a^b f(x)g(x)\, d\alpha$$

converges.

Homework Equations

The Attempt at a Solution

I've tried putting a uniform bound M on F, and finding $x_0$ such that $x_0 < x < b$ implies that $0 \leq g(x) < \frac{\epsilon}{M}$. However, I can't find a way to bring $g(x)$ "inside" the integral, so to speak. My intuition is that at worst $F$ is oscillating and that multiplying by g will bring it under control to make it convergent, sort of like $\int_1^\infty \frac{\sin x}{x} dx$ but I just can't get any ideas to work, and I'm really stuck.

 

[mighty2000]

Thank you for your post. I can provide some guidance and suggestions to help you solve this problem.

Firstly, it is important to understand the definitions and properties of the functions involved in this problem. For example, monotone increasing functions are functions that either increase or stay constant as the input increases. Similarly, monotone decreasing functions are functions that either decrease or stay constant as the input increases. It may be helpful to draw a graph of these functions to get a better understanding of their behavior.

Next, it is important to understand the meaning of integrability. A function is said to be integrable over an interval if its integral (area under the curve) exists and is finite. This means that the function is well-behaved and does not have any extreme or erratic behavior over that interval.

To prove that the given integral converges, you can use the definition of convergence, which states that a series or integral converges if the limit of its terms or integrands approaches a finite value. In this case, we can use the definition of convergence to prove that the given integral converges.

To start, you can use the fact that F(x) is bounded and g(x) is monotone decreasing to find a uniform bound for the product f(x)g(x). This will help you establish that the integral is bounded and thus converges.

Next, you can use the fact that g(x) is non-negative and converges to 0 as x approaches b from the left to show that the product f(x)g(x) approaches 0 as x approaches b from the left. This will help you establish that the integral converges to 0.

Finally, you can use the monotonicity of g(x) to show that the integral is decreasing as x approaches b from the left. This will help you establish that the integral converges to a finite value.

I hope these suggestions are helpful in guiding you towards a solution. Remember to always carefully define and understand the functions and properties involved, and use the definitions and properties to guide your proof. Good luck!