MHB Real Analysis - Riemann Integral Proof

[joypav]

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I have no idea how to incorporate the limit into the basic definitions for a Riemann integral? All we have learned so far is how to define a Riemann integral and the properties of Riemann integrals. What should I be using for this?

 

[Euge]

Welcome, joypav! (Wave)

Without loss of generality, assume $f(0) = 0$. For the result is equivalent to $\lim\limits_{n\to \infty} \int_0^1 g(x^n)\, dx = 0$, where $g(t) = f(t) - f(0)$. Let $\epsilon > 0$. By continuity of $f$ at $0$, there is a positive number $\delta\in (0,1)$ such that for all $t\in [0,1]$, $\lvert t\vert < \delta$ implies $\lvert f(t)\rvert < \epsilon$. Now write

$$\int_0^1 f(x^n)\, dx = \int_0^{\sqrt[n]{\delta}} f(x^n)\, dx + \int_{\sqrt[n]{\delta}}^1 f(x^n)\, dx$$

Since $\lvert f\rvert$ is continuous on $[0,1]$, it has a maximum value, $M$. Show that $\int_{\sqrt[n]{\delta}}^1\, f(x^n)\, dx$ is bounded by $M(1 - \sqrt[n]{\delta})$, and that $\int_0^{\sqrt[n]{\delta}} f(x^n)\, dx$ is bounded by $\epsilon \sqrt[n]{\delta}$. Then

$$\left\lvert \int_0^1 f(x^n)\, dx \right\rvert \le \epsilon \sqrt[n]{\delta} + M(1 - \sqrt[n]{\delta})$$

Letting $\epsilon \to 0^+$, we obtain

$$\left\lvert \int_0^1 f(x^n)\, dx\right\rvert \le M(1 - \sqrt[n]{\delta})$$

Finish the argument using the squeeze theorem.

 

[joypav]

I see. So it relies mostly on the continuity of f on the interval. We can prove the bounds using that the integral <= M(b-a).

I do have a question though.. why do we not bound the integral by epsilon(1-(delta)^1/n) and epsilon(delta)^1/n. Then when you add up the two integrals you would get that the whole integral from 0 to 1 is <= 2epsilon. And because of continuity, we can make this 0 by letting epsilon approach 0.

I guess my question is, why do we use M at all? Doesn't epsilon bound f(t)?

(Sorry, I'm typing on my phone, so can't use the correct symbols.)

 

[Euge]

We cannot bound the second integral by $\epsilon (1-\delta^{1/n})$, because it is not true that $\lvert f(t)\rvert < \epsilon $ whenever $\lvert t\rvert \ge \delta $. If we had that, then there would be no need to break up the integral into two parts, as we've done here.

 

[joypav]

We cannot bound the second integral by $\epsilon (1-\delta^{1/n})$, because it is not true that $\lvert f(t)\rvert < \epsilon $ whenever $\lvert t\rvert \ge \delta $. If we had that, then there would be no need to break up the integral into two parts, as we've done here.

I see. Thank you