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

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.

• Note that $f$ is Lebesgue-integrable, then the gauge integral is the same that de Lebesgue integral. – Martín Vacas Vignolo Dec 23 '17 at 20:33

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 $$\delta_2$$.

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 $0$

• Thank you for answering, but I actually havent been taught what Lebesgue integral, Lebesgue mesure or Lebesgue-integrable means... :( – Digitalis Dec 23 '17 at 20:44