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Before I start, I know no measure theory. All I know is the following definition:

A set $A \subset \Bbb{R}^n$ is said to have measure zero, if for every $\varepsilon > 0$ there exits rectangles $\{R_k\}_{k=1}^ \infty$ such that $A \subset \bigcup_{k=1}^\infty R_k$ and $\sum_{k=1}^\infty \mu(R_k) < \varepsilon$.

Now here's my question: If $A$ is a set of measure zero and $f$ is a Riemann integrable functions then its Riemann integral over $A$ , $\int_A f$ is zero.

This what I tried:(I'm trying to show the integral is less than epsilon) Let the $R_k$'s be the ones in the definition then we have $$\left|\int_A f\right| \le \left|\int_{\bigcup_{k=1}^\infty R_k} f \right|$$ I'm doubting this.

Then $|\int_{\bigcup_{k=1}^\infty R_k} f |\le |\sup f| \sum_{k=1}^\infty \mu(R_k) $ , not exactly what I wanted :/

I know this question is similar to this one but I don't want to use the convention in the first answer and the second answer uses things which I don't know.

EDIT : According to nulluser the question " is true if $\int$ is the Lebesgue integral. Alternatively, the theorem is also true if you assume $A$ has Peano-Jordan measure zero (i.e. take a finite collection $\{R_k\}_{1}^n$ instead of a countable one in the definition above). In this case $1_Af$ is guaranteed to be Riemann integrable. Alternatively you could assume $1_Af$ is Riemann integrable and try to proceed."

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  • You have exactly what you need, you are just confused because you used epsilon twice instead of a new variable for the bound of $\int_A f$. – vadim123 May 27 '13 at 15:24
  • Why is it not what you wanted? Just use the inequality from the definition, and then consider that it is true for each $\epsilon>0$. – celtschk May 27 '13 at 15:25
  • Oh!! So I pick the first epsilon in the definition to be $\varepsilon/ |\sup f|$ ? – user63697 May 27 '13 at 15:26
  • Is there any positive value you cannot reach with $\left|\sup f\right|\epsilon$ when $\epsilon$ can be an arbitrary positive number? – celtschk May 27 '13 at 15:28
  • OK, I see one possible problem now: The supremum might be infinity. In that case, you might divide the set into countably many subsets which have a finite supremum each. – celtschk May 27 '13 at 15:30
  • I assumed the function is bounded since in all the times I've studies integrals the functions were assumed bounded – user63697 May 27 '13 at 15:31
  • Yes. One only defines Riemann integrals of bounded functions. (Otherwise, we're in the land of improper integrals.) – Ted Shifrin May 27 '13 at 15:34
  • I am starting to feel in doubt of the big inequality, does any one have a proof of it? – user63697 May 27 '13 at 15:51
  • This question is ill posed. There are Lebesgue measurable sets with measure zero which are not Peano-Jordan measurable. If you aren't using the Lebesgue integral then there is no guarantee that $\int_A f$ even makes sense. – nullUser May 27 '13 at 16:35
  • @nullUser I believe I should add $f$ is Riemann integrable , why wouldn't the integral make sense? – user63697 May 27 '13 at 16:43
  • @user63697 Suppose $f(x)=1$ on $[0,1]$ and I choose $A = \mathbb{Q} \cap [0,1]$. You have probably seen that $\lambda(\mathbb{Q})=0$ since $\mathbb{Q}$ is countable. But the function $\chi_{\mathbb{Q}}$ is not Riemann integrable on $[0,1]$. – nullUser May 27 '13 at 16:46
  • @nullUser These notations are new to me I'm guessing by $\lambda (\Bbb{Q})$ you mean the measure of $\Bbb{Q}$ which is zero ( I believe it can be showed by using the fact $\Bbb{Q}$ is countable), but what is $\chi_{\mathbb{Q}}$ ? – user63697 May 27 '13 at 16:55
  • @user63697 Yes, I am using $\lambda(\mathbb{Q})$ to mean the Lebesgue measure on $\mathbb{R}$, the definition of "measure zero" above coincides with having Lebesgue measure zero. The notation $\chi_\mathbb{Q} = 1_\mathbb{Q}$ stands for the indicator function of $\mathbb{Q}$, i.e. $\chi_\mathbb{Q}(x)=1$ if $x \in \mathbb{Q}$ and $\chi_\mathbb{Q}=0$ otherwise. – nullUser May 27 '13 at 16:58
  • @nullUser I don't understand the relation between $f$ and $\chi_{\mathbb{Q}}$, doesn't $f$ remain integrable over $A$ ? – user63697 May 27 '13 at 17:05
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    @user63697 $\int_A f$ is defined to be $\int f \chi_A$. This means that you need $f \chi_A$ to be Riemann integrable. In our case $f \chi_A = 1 \cdot \chi_{\mathbb{Q}} = \chi_{\mathbb{Q}}$. – nullUser May 27 '13 at 17:10
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    @nullUser These are new to me but very interesting. I got it now, so it would be better to use the Lebesgue integral here. – user63697 May 27 '13 at 17:13
  • @user63697 Yes, the theorem is true if $\int$ is the Lebesgue integral. Alternatively, the theorem is also true if you assume $A$ has Peano-Jordan measure zero (i.e. take a finite collection ${ R_k }_1^n$ instead of a countable one in the definition above). In this case $1_A f$ is guaranteed to be Riemann integrable. Alternatively you could assume $1_A f$ is Riemann integrable and try to proceed. – nullUser May 27 '13 at 17:17
  • I'll edit this in. – user63697 May 27 '13 at 17:19

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