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Math Help - Analysis, riemann and lebesque integrable

  1. #1
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    Analysis, riemann and lebesque integrable

    Denote by m the Lebesgue measure on X = (0, 1).
    d(\cdot) is the Dirichlet function.


    Consider the sequence of functions on (0, 1) defined as follows: f_N(x)=\left\{\begin{array}{cc}0,&\mbox{ if }<br />
x\mbox{ is irrational or }x = \frac{k}{n}\mbox{ (in lowest terms) and }n>N\\1, & \mbox{ if }x = \frac{k}{n}\mbox{ (in lowest terms) and }n \leq N\end{array}\right.

    1) Prove that f_N(x) < f_{N+1}(x), \  x \in (0, 1), and that \lim_{N\rightarrow \infty} f_N(x) = d(x), \ x \in<br />
(0, 1).
    2) Are the functions f_N Riemann integrable? If your answer is yes, what is  \int_0^1 f_N (x)dx?
    3) Are the functions f_N Lebesgue integrable? If your answer is yes, what is the Lebesgue integral \int f_N(x)dm?
    4) Is it true that \lim \int_0^1 f_N(x)dx = \int_0^1 d(x)dx (the Riemann integral)?
    5). Is it true that \lim \int_0^1 f_N(x)dx \leq \int d(x)dm (the Lebesgue integral)?
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  2. #2
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    Quote Originally Posted by JJMC89 View Post
    Denote by m the Lebesgue measure on X = (0, 1).
    d(\cdot) is the Dirichlet function.

    Consider the sequence of functions on (0, 1) defined as follows: f_N(x)=\left\{\begin{array}{cc}0,&\mbox{ if }<br />
x\mbox{ is irrational or }x = \frac{k}{n}\mbox{ (in lowest terms) and }n>N\\1, & \mbox{ if }x = \frac{k}{n}\mbox{ (in lowest terms) and }n \leq N\end{array}\right.

    1) Prove that f_N(x) < f_{N+1}(x), \  x \in (0, 1), and that \lim_{N\rightarrow \infty} f_N(x) = d(x), \ x \in<br />
(0, 1).
    2) Are the functions f_N Riemann integrable? If your answer is yes, what is  \int_0^1 f_N (x)dx?
    3) Are the functions f_N Lebesgue integrable? If your answer is yes, what is the Lebesgue integral \int f_N(x)dm?
    4) Is it true that \lim \int_0^1 f_N(x)dx = \int_0^1 d(x)dx (the Riemann integral)?
    5). Is it true that \lim \int_0^1 f_N(x)dx \leq \int d(x)dm (the Lebesgue integral)?

    This may not be much help but I can tell you that the Lebesgue integral exists and is 0. This is because the function is equal to 0 a.e. (since the rationals are a set of measure 0).
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