Pointwise converging function, with integral divergent

Question

I tried to make a trigonometric function looking something like this

$f_n = \frac{sin(nx)}{n}$

but I can't get it to diverge after the integration. All help appreciated (with both (a) and (b))

Answer 1

I would suggest a different approach, which to me seems simple. The simplest functions to integrate are step functions:

$$ g(x)= c\cdot\mathbf{1}_{I}(x) \quad \text{where} \quad I\subseteq [0,1] \; \text{is an interval} $$

The integral is:

$$ \int_0^1 g(x)dx= c\cdot \vert I\vert \quad \text{where} \; \vert I\vert \text{ is the length of } I $$

So for example for (a), if you have a sequence of functions:

$$ f_n(x)= a_n\cdot \mathbf{1}_{ (0,\frac{1}{n}] }(x) $$

Their integrals would be:

$$ \int_0^1 f_n(x)dx= a_n \cdot \Big( \frac{1}{n}-0 \Big)=\frac{a_n}{n} $$

You can verify that they converge pointwise to $0$ and are integrable. Choosing $a_n$ appropriately will give you your desired counter-examples.