Analysis, riemann and lebesque integrable

**JJMC89** · April 27th 2010, 01:29 PM

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)?

**southprkfan1** · April 27th 2010, 03:01 PM

Originally Posted by JJMC89

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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