Show that the following function has a gauge-integral and calculate its value.

Question

We define:

$$f: I = [0,1] \rightarrow \mathbb{R}: x \mapsto f(x) = \sin(x)1_{[0,1] - \mathbb{Q}} + x1_{[0,1]\cap\mathbb{Q}} $$

where $1_A$ is the characteristic function on A.

Show that $f$ has a gauge integral on [0,1] and calculate it.

I've come across this problem and I don't have a clue on how to solve it. I thought about using the definition of gauge integral but that would require me to "guess" what the value of $\int_I f$ actually is. Having no clue what $\int_I f$ is worth I've been thinking about using Cauchy's criterion instead but without much success.

Any help or clues appreciated.

Answer 1

It's easily to show gauge-integrability using one math trick.

All the folowing will be done in assumption that gauge-integral is Henstock-Kurzweil integral. (H.-K. integral).

We know that if $f$ and $g$ are H.-K. integrable then $f+g$ also H.-K. integrable.

So, let's rewrite your function as

$$ \sin(x) + (x-\sin(x))1_{[0,1]\cap\mathbb{Q}} $$

First part, $f = \sin(x)$ is Riemann integrable (and consequently H.-K. integrable with gauge $\delta_1(x)$ equals some constant value, depending on $\varepsilon$).

Second part, $g = (x-sin(x))1_{[0,1]\cap\mathbb{Q}}$ is H.-K. integrable with gauge $$\delta_2(x) = \bigg(\frac{\varepsilon}{2^{k+1}}\bigg)1_{[0,1]\cap\mathbb{Q}} + 1_{[0,1]\setminus\mathbb{Q}}$$

where $k$ means number of rational number.

For any $\tau=\{\xi_i,\Delta_i\}_{i=1}^n$ - $\delta_2$-partition of $[0,1]$ sum $$\bigg|\sum_{i=1}^n g(\xi_i)|\Delta_i|\bigg| \leq \sum_{k=1}^\infty g(r_k)2\delta(r_k) < \sum_{k=1}^\infty \frac{2\varepsilon}{2^{k+1}} = \varepsilon$$ where $r_k$ means $k$-th rational number. That's why $\int_{[0,1]}g(x)dx = 0$

So, your function is integrable as sum of integrable functions. If you want an explicit formula for gauge, you can try to construct it from $\delta_1$ and $\delta_2$.

Answer 2

Because $f$ is Lebesgue-integrable, then the gauge integral is the same that the Lebesgue integral. Now, for this:

$\int_I f=\int_I sin(x)=-cos(1)+1$

The first $=$ works because the Lebesgue integral over a $0$-measure set is $0$