Question

Let C is a subset of [0,1] be the cantor set and let f: [0,1]→[0,∞) be given by f(x)=0 on C and f(x)=n, in each complementary interval of length 3^(-n) show that f is lebesgue measurable and compute ∫_0^1▒〖f(x)〗dx

Answer 1

Answer on Question #54156 – Math – Real Analysis

Question

Let $C$ be a subset of $[0,1]$, the cantor set and let $f\colon [0,1]\to [0,\infty)$ be given by $f(x) = 0$ on $C$ and $f(x) = n$ in each complementary interval of length $3^{-n}$. Show that $f$ is Lebesgue measurable and compute $\int_0^1 f(x)dx$.

Solution

$$f(x) = \begin{cases} 0 & \text{on } C \\ n & \text{on } [0,1] \setminus C \end{cases} = 0 \cdot \chi_C(x) + n \cdot \chi_{[0,1] \setminus C}(x) = \sum_{k=1}^{n} a_k \chi_{A_k}, \, x \in [0,1],$$

where $\chi_C(x) = \begin{cases} 1 & \text{if } x \in C, \\ 0, & \text{if } x \notin C, \end{cases}$ is a characteristic function of a set $C$,

$$\chi_{[0,1] \setminus C}(x) = \begin{cases} 1 & \text{if } x \in [0,1] \setminus C, \\ 0, & \text{if } x \notin [0,1] \setminus C, \end{cases} \text{ is a characteristic function of a set } [0,1] \setminus C.$$

The characteristic function of a set $E$ is measurable if and only if $E$ is measurable.

The cantor set $C$ is a Borel set and hence measurable. The closed interval $[0,1]$ is a Borel set and hence measurable.

Set $[0,1] \setminus C$ is Borel and hence measurable as complement of a measurable set $C$.

Thus, functions $\chi_C(x), \chi_{[0,1] \setminus C}(x)$ are Lebesgue measurable, because sets $C$ and $[0,1] \setminus C$ are measurable.

By properties of measurable functions, functions $0 \cdot \chi_C(x)$ and $n \cdot \chi_{[0,1] \setminus C}(x)$ will be Lebesgue measurable, besides, function $f(x) = 0 \cdot \chi_C(x) + n \cdot \chi_{[0,1] \setminus C}(x)$ is Lebesgue measurable as the sum of two Lebesgue measurable functions.

We showed that function $f$ is Lebesgue measurable.

Next, function $f(x) = 0 \cdot \chi_C(x) + n \cdot \chi_{[0,1] \setminus C}(x)$ is simple, therefore

$$\int_0^1 f(x) \, dx = \sum_{k=1}^{n} a_k \mu(A_k) = 0 \cdot \mu(C) + n \cdot \mu([0,1] \setminus C) = 0 + n = n,$$

because

$$\mu(C) = 0, \, \mu([0,1]) = 1 - 0 = 1.$$

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