Real Analysis - Prove the Riemann Integral Converges

[joypav]

Just a couple questions.

Problem 2: Just would like to know if this is the correct approach for this problem.

Problem 3: I am just wondering if I can use Problem 2 to prove the first part of Problem 3? Because to me, they seem very similar.

Problem 4: Would I use the MVT for integrals here?

It's not necessary for me to get a full answer, just wanted to make sure I was starting them correctly.

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Problem 2: Just would like to know if this is the correct approach for this problem.

How do you know that the positive and negative parts of $f$ are improper Riemann integrable? There needs to be more justification.

A way to avoid the issue is to consider the inequality

$$\left\lvert \int_a^c f(x)\, dx - \int_a^d f(x)\, dx\right\rvert \le \int_d^\infty \lvert f(x)\rvert\, dx$$

for $c > d$.

Problem 3: I am just wondering if I can use Problem 2 to prove the first part of Problem 3? Because to me, they seem very similar.

Yes, you can.

Problem 4: Would I use the MVT for integrals here?

No. You'll need to prove that $f(c+)$ and $f(c-)$ exist (unless you're allowed to assume that) -- that's where monotonicity of $f$ is used. An $\epsilon-\delta$ argument will do for this problem.

 

[joypav]

No. You'll need to prove that $f(c+)$ and $f(c-)$ exist (unless you're allowed to assume that) -- that's where monotonicity of $f$ is used. An $\epsilon-\delta$ argument will do for this problem.

I see. $f$ is monotone increasing on the closed interval $[a,b]$, so it is bounded above. Then show that $f(c+)$ is equal to $inf(f(x))$ where $c<x$. Similarly for $f(c-)$.

I can also assume, by the definition given for $F(x)$, that $f\in R[a,b]$ and $F'(c)=f(c)$ for all $c\in[a,b]$? Then I can use these assumptions to rewrite $F_+^{'}(c)$ and show it is equal to $f(c+)$?

 

[joypav]

How do you know that the positive and negative parts of $f$ are improper Riemann integrable? There needs to be more justification.

A way to avoid the issue is to consider the inequality

$$\left\lvert \int_a^c f(x)\, dx - \int_a^d f(x)\, dx\right\rvert \le \int_d^\infty \lvert f(x)\rvert\, dx$$

for $c > d$.

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Can we not use the above argument to show that they are Riemann integrable? Using the fact that f is absolutely integrable.

 

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I can also assume, by the definition given for $F(x)$, that $f\in R[a,b]$ and $F'(c)=f(c)$ for all $c\in[a,b]$? Then I can use these assumptions to rewrite $F_+^{'}(c)$ and show it is equal to $f(c+)$?

No, but one argues that a monotone function on a closed interval $[a,b]$ belongs to $R[a,b]$. So then $F$ would make sense. Use an $\epsilon-\delta$ argument to show $F_+^{'}(c) = f(c+)$ and similarly for $F_{-}^{'}(c)$.

Can we not use the above argument to show that they are Riemann integrable? Using the fact that f is absolutely integrable.

No. For the indicator function on the rationals (also known as Dirichlet's function) is bounded by Riemann integrable functions, but it is not itself Riemann integrable. Typically, to prove Riemann integrability, partitions or tagged partitions are used.

 

[joypav]

No. For the indicator function on the rationals (also known as Dirichlet's function) is bounded by Riemann integrable functions, but it is not itself Riemann integrable. Typically, to prove Riemann integrability, partitions or tagged partitions are used.

I see. Being bounded does not prove that it is Riemann integrable. Thank you for the help. My professor is a stickler for detail, and rightfully so.

 

[joypav]

No, but one argues that a monotone function on a closed interval $[a,b]$ belongs to $R[a,b]$. So then $F$ would make sense. Use an $\epsilon-\delta$ argument to show $F_+^{'}(c) = f(c+)$ and similarly for $F_{-}^{'}(c)$.

I'm sorry, but I just can't figure out this problem. I finished the other one, but I'm still stuck on this question. Do you have any other advice?

 

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Yes. Let $h > 0$ such that $c + h\in (a,b)$. Then

$$F(c + h) - F(c) - f(c+)h = \int_c^{c+h} [f(t) - f(c+)]\, dt $$

Let $\epsilon > 0$. There exists a $\delta > 0$ such that for all $t$, $c < t < c + h$ implies $\lvert f(t) - f(c+)\rvert < \epsilon$. Now prove $\lvert F(c + h) - F(c) - f(c+)h\rvert < \epsilon h$ whenever $0 < h < \delta$. Then $F_+'(c) = f(c+)$. Argue similarly to show $F_{-}'(c) = f(c-)$.