# Example of a function on $([0,1]; \sigma ([0;1]); \lambda )$ for which it has no meaning to write $\int f d \lambda$

I am trying to understand Lebesgue integration and in order to understand well this concept I would like to have an example of a function $$(0,1]$$ ( or $$[0,1) , [0,1] , (0,1)$$ ) for which it has no meaning to write $$\int f d \lambda$$ with $$([0,1]; \sigma ([0,1]); \lambda )$$ with $$\lambda =$$ Lebesgue measure.

For example understanding why the Dirichlet function is not Riemann integrable but it is Lebesgue integrable helped me a lot to understand the concept of Lebesgue integrable. But I am a little bit stuck in my understanding because now it seems to me that all the functions can be Lebesgue integrable as the Lebesgue integration authorize the integral to be equal to $$\infty$$. Where it is not the case for the Riemann definition of integration which impose the convergence of the sum (thus has to be finite).

I thought about the function $$\frac{-1}{x}$$ as not been Lebesgue integrable on $$([0,1]; \sigma ([0,1]); \lambda )$$ but I am not sure that this is correct and more precisely I can not justify it rigorously.

Any help precision/correction in what I have written above and details comment will be cool and greatly appreciated!

• RI functions are also LI with the same value for the intergal, so $\int_a^{1}\frac 1 x dx=-\ln a$. Let $a \to 0+$. Feb 25 at 12:13
• The name is Lebesgue Feb 25 at 12:39
• Riemann integrable and Lebesgue integrable. Feb 25 at 12:41
• What's wrong with the $1/x$ example @geetha290krm provided above? Feb 25 at 16:18
• @OffHakhol Yes when $f$ is positive measurable (so $f(x) \geq 0$ for all $x \in \Omega$, the domain that we consider) and $\int f d \lambda = \infty$ one usually says $f$ is not integrable. Feb 25 at 18:39

If you allow the Lebesgue integral to have infinite value, I deduce your convention for Lebesgue integrability is when $$\int f_+dx,\int f_- dx$$ are not both infinite. There are a few nice example to help your intuition about Riemann and Lebesgue integrability:

• Riemann and Lebesgue integrals have the same value for continuous functions on compact intervals

• Some functions are Riemann integrable but not Lebesgue integrable, this is the case for say $$f(x)=\sin(x)/x$$. The limit definition of the Riemann integral will converge, while the integrals of $$f_+,f_-$$ are both infinite. Another nice example is the Cauchy base 13 function, as the support has measure zero.

• Most Lebesgue measurable functions are not Riemann Integrable. This is the case for some classical examples such as $$f(x)=\mathbb{1}_{\mathbb{Q}\;\cap\; [0,1]}$$.

• Many functions are not Lebesgue integrable. One can easily construct a function such that both $$f_-,f_+$$ have infinite integrals.

Regarding $$f(x)=1/x$$ on say $$(0,1]$$, it can be seen as Lebesgue integrable if you allow the Lebesgue integral to have infinite value. However the most common convention is, I belive, to require $$\int |f(x)|dx<\infty$$

• thk a lot for your answer i am going to read it Feb 28 at 11:41
• So is there a function for which $$\int f d \lambda$$ will have no meaning? Feb 28 at 11:44
• yes like $sin(x)$ over $\mathbb{R}$, where both the positive and negative parts have infinite integrals. Feb 28 at 15:08
• thank a lot. But $|sin(x)|$ not right? Feb 28 at 15:19
• For $|\sin(x)|$, it depends on your definition, some people may allow the Lebesgue integral to be infinite, but what is clear is that you can't have $\int f= \int f_+-\int f_-$ if both $f_+$ and $f_-$ have infinite integrals, giving easy examples of functions that are not lebesgue integrable Feb 28 at 15:43