If A is dense in [0,1] and f(x) = 0, x in A, prove ∫fdx = 0.

[Eclair_de_XII]

Homework Statement

"A set $A\subset [0,1]$ is dense in $[0,1]$ iff every open interval that intersects $[0,1]$ contains $x\in A$. Suppose $f:[0,1]\rightarrow ℝ$ is integrable and $f(x) = 0,x\in A$ with $A$ dense in $[0,1]$. Show that $\int_{0}^{1}f(x)dx=0$."

Homework Equations

Let $P=\{x_1,...,x_n\}$ be a partition of $[0,1]$

$M_i=sup\{f(x):x\in[x_{i-1},x_i]\}$
$m_i=inf\{f(x):x\in[x_{i-1},x_i]\}$

$U(P,f)=\sum_{i=1}^nM_i(f)(x_i-x_{i-1})$
$L(P,f)=\sum_{i=1}^nm_i(f)(x_i-x_{i-1})$

$\int_{0}^{1} fdx=sup\{L(P,f):P \space \text {a partition of} \space [0,1]\}=inf\{U(P,f):P \space \text {a partition of} \space [0,1]\}$

The Attempt at a Solution

I haven't actually attempted a solution yet; everything that follows is scratchwork...

Basically, I think I have to use the fact that $A$ being dense in $[0,1]$ gives the following implication:

$\int_{x\in A}f(x)dx=0 \space ⇒ \int_{0}^{1} f(x)dx=0$,

since $f(x)=0,\forall x\in A$, then $\int_{x\in A} f(x)dx=0$.

So my plan is to construct a series of open intervals that intersect $[0,1]$ as such:

$Q_i=(a_i,b_i)$ where $a_i=b_{i-1}$

Then, I define the initial open interval as $Q_0=(a_0,b_0)$, with $a_0\leq 0$ and $b_0 > 0$ and the final open interval as $Q_n=(a_n,b_n)$ where $b_n\geq 1$.

Then I construct a partition using those open intervals with $(x_{i-1},x_i)=(a_i,b_i)$, with $x_0=0$ and $x_n=1$. So I would construct a lower and upper sum, which by the conditions of the problem, must be equal to each other, and equal to zero (still going to need to figure this part out).

This is my current idea of how to handle the problem. Can anyone tell me if anything I have written is off? Thanks.

 

[andrewkirk]

The claim is not true with the usual definition of integrable, which is Lebesgue-integrable. The Indicator function that is 1 on the irrational numbers and zero elsewhere is Lebesgue-integrable and has integral 1 on [0,1], even though the set on which it is zero - the rational numbers in [0,1] - is dense.

The claim may be true if we restrict ourselves to Riemann integration, as the above indicator function is not Riemann-integrable.

From your plan outline, it looks like you are working with Riemann integrals. Some element of the proof will have to exclude cases like the irrational indicator function on the grounds that they are not Riemann-integrable. Also, the proof will need to use the density of $A$, a fact that the above plan does not reference.

I think the proof can be easier than your plan.
First prove that the 'sup' bit is zero, using the density premise. This is the only bit where you need to refer to partitions.
Then complete the proof using the premise that $f$ is integrable. What does that premise tell us about the relation between the 'sup' bit and the 'inf' bit?

 

[Eclair_de_XII]

The claim may be true if we restrict ourselves to Riemann integration, as the above indicator function is not Riemann-integrable.

Sorry; I should have mentioned that this class is working with Riemann integration, and that $f$ is Riemann-integrable.

First prove that the 'sup' bit is zero, using the density premise. This is the only bit where you need to refer to partitions.

My guess:

Let $O$ be an open interval such that $O\cap [0,1]\neq ∅$.

Because $A$ is dense in $[0,1]$, $O\cap [0,1] \subseteq A$. So let $P=\{x_1,...,x_n\}$.

Then, since $A$ is dense in $[0,1]$ and if $x\in (a,b)\subset [0,1]$, then $f(x)=0$.

So $m_i=inf\{f(x):x\in (x_{i-1},x_i)\subset [x_{i-1},x_i]\}=0,\forall i\in ℕ$, since we assume that $f(x)\geq 0,\forall x\in ℝ$.

So $sup\{L(P,f)\}=0$. Since $f$ is Riemann-integrable, $sup\{L(P,f)\}=inf\{U(P,f)\}=\int_{0}^{1} f(x)dx=0$.