How to show that $\chi_{[0,\infty)}(x)$is not integrable?

Question

How to show that $\chi_{[0,\infty)}(x)$is not integrable?

My idea:

to use the following proposition:

Let the nonnegative function $f$ be integrable over $E.$ then $f$ is finite a.e. on $E.$

But I am not sure, I feel that my function is finite even though its domain is infinite. Any help will be appreciated.

Answer 1

1. Show by definition that for any given positive integer $n$, $\chi_{[0,\infty)}(x)\ge \chi_{[0,n]}(x)$ for all $x$.

2. Show by Theorem 10 in Royden that for each $n$.

   $$ \int \chi_{[0,\infty)}\ge \int \chi_{[0,n]} $$

3. Show by Definition in Section 4.2 that for each positive integer $n$, $$ \int \chi_{[0,n]} = 1\cdot m([0,n])=n $$ Note that $\chi_{[0,n]}$ is a simple function.

4. Combine 1--3 to show that $\int \chi_{[0,\infty)} =\infty$ and thus $\chi_{[0,\infty)}$ is not integrable.

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Note: bounded measurable functions are NOT necessarily Lebesgue integrable. You have just seen an example. But bounded measurable functions over a set of finite measure are.