# dense

• Mar 8th 2009, 11:52 PM
manjohn12
dense
Let $D$ be a dense subset of $[a,b]$. Suppose that $f$ is integrable on $[a,b]$ and $f(x) = 0$ for all $x \in D$. Prove that $\int_{a}^{b} f = 0$.

So every point in $[a,b]$ is either is in $D$ or a limit point of $D$. So define $g(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \in D \\ f(x) \ \text{if} \ x \notin D \end{cases}$.

From here how would we show that $\int_{a}^{b} f = 0$?
• Mar 9th 2009, 03:38 AM
Opalg
Quote:

Originally Posted by manjohn12
Let $D$ be a dense subset of $[a,b]$. Suppose that $f$ is integrable on $[a,b]$ and $f(x) = 0$ for all $x \in D$. Prove that $\int_{a}^{b} f = 0$.

So every point in $[a,b]$ is either is in $D$ or a limit point of $D$. So define $g(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \in D \\ f(x) \ \text{if} \ x \notin D \end{cases}$.

From here how would we show that $\int_{a}^{b} f = 0$?

(This result is false for the Lebesgue integral, so I'll assume that it refers to the Riemann integral.)

Show that, for any dissection of the interval [a,b], the lower Riemann sum for f is ≤0 and the upper Riemann sum is ≥0. It follows that if f is integrable then its integral is 0.
• Mar 12th 2009, 11:54 AM
manjohn12
Quote:

Originally Posted by Opalg
(This result is false for the Lebesgue integral, so I'll assume that it refers to the Riemann integral.)

Show that, for any dissection of the interval [a,b], the lower Riemann sum for f is ≤0 and the upper Riemann sum is ≥0. It follows that if f is integrable then its integral is 0.

But suppose that you did not have the notion of upper and lower Riemann sums?

Proof. Let $\epsilon >0$. Choose $\delta < \frac{\epsilon}{2}$. Let $P = \{x_0, x_1, x_2, \ldots, x_n \}$ be a partition of $[a,b]$ with $||P|| < \delta$. Choose $x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}$ arbitrarily such that $x_{i-1} \leq x_{i}^{*} \leq x_{i}$. We know that $\mathcal{R}(f,P) = \sum_{i=1}^{n} f(x_{i}^{*})(x_{i}-x_{i-1})$. Now $\mathcal{R}(f,P) = 0$ if $x_{i}^{*} \in D$. If $x_{i}^{*} \notin D$, then it is a limit point of $D$. From here how would you conclude that $\int_{a}^{b} f = 0$?
• Mar 12th 2009, 02:56 PM
HallsofIvy
Quote:

Originally Posted by manjohn12
Let $D$ be a dense subset of $[a,b]$. Suppose that $f$ is integrable on $[a,b]$ and $f(x) = 0$ for all $x \in D$. Prove that $\int_{a}^{b} f = 0$.

So every point in $[a,b]$ is either is in $D$ or a limit point of $D$. So define $g(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \in D \\ f(x) \ \text{if} \ x \notin D \end{cases}$.

There is no point in doing that: because you are told that f(x)= 0 for all x in D, your g(x) is f(x).

Quote:

From here how would we show that $\int_{a}^{b} f = 0$?
• Mar 12th 2009, 03:02 PM
HallsofIvy
Quote:

Originally Posted by manjohn12
But suppose that you did not have the notion of upper and lower Riemann sums?

Proof. Let $\epsilon >0$. Choose $\delta < \frac{\epsilon}{2}$. Let $P = \{x_0, x_1, x_2, \ldots, x_n \}$ be a partition of $[a,b]$ with $||P|| < \delta$. Choose $x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}$ arbitrarily such that $x_{i-1} \leq x_{i}^{*} \leq x_{i}$. We know that $\mathcal{R}(f,P) = \sum_{i=1}^{n} f(x_{i}^{*})(x_{i}-x_{i-1})$. Now $\mathcal{R}(f,P) = 0$ if $x_{i}^{*} \in D$. If $x_{i}^{*} \notin D$, then it is a limit point of $D$. From here how would you conclude that $\int_{a}^{b} f = 0$?

That is the "lower Riemann sum" Opalg is talking about! In any interval of any partition, you can always choose an $x^*$ so that $f(x^*)= 0$ so you can always have $\mathcal{R}(f,P)= 0$. Now, because you are given that the function is integrable, the limit, using any finer and finer partitions must be the same.
• Mar 12th 2009, 03:33 PM
manjohn12
Quote:

Originally Posted by HallsofIvy
That is the "lower Riemann sum" Opalg is talking about! In any interval of any partition, you can always choose an $x^*$ so that $f(x^*)= 0$ so you can always have $\mathcal{R}(f,P)= 0$. Now, because you are given that the function is integrable, the limit, using any finer and finer partitions must be the same.

So what does denseness have to do with this problem? Is is just extra information?
• Mar 13th 2009, 12:45 AM
Opalg
Quote:

Originally Posted by manjohn12
So what does denseness have to do with this problem? Is is just extra information?

You need the denseness of D to conclude that every interval of a partition contains a point x* (an element of D) for which f(x*)=0.