$f: [0,1] \rightarrow \mathbb{R}$ non-decreasing. Show $f \in L^1.$

Asked Jan 08 '21 at 18:00

Active Jan 08 '21 at 18:00

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I read the proof of the following statement:

$f: [0,1] \rightarrow \mathbb{R}$ non-decreasing. Then $f \in L^1.$ Where $L^1$ denotes the set of lebesgue-integrable functions.

Proof:

Let $N$ be the set of discontinuous points of $f$. At every point of $x \in N$ there exists due to monotonicity of $f$ both left and right limit $L_x, R_x, L_x \neq R_x$. Then we choose a rational number $q_x$ s.t. $L_x < q_x < R_x$. Clearly if $x< y$ it follows that $q_x < q_y$. Hence, there is an injection from $N$ to $\mathbb{Q}$. Hence, $N$ is countable and has Lebesgue-measure $0 \implies f\in L^1$.

The point I don't understand is, why is it enough to have an injection from $N$ to $\mathbb{Q}$? We do need a surjective function as well, right?

asked Jan 08 '21 at 18:00

Well, no. N could be finite or even empty. So it can be a lot "smaller" than Q. – Andrew Kay Jan 08 '21 at 18:06
Q is just being used to keep the discontinuous points in order. – Andrew Kay Jan 08 '21 at 18:09
Ah, and if I have to sets $A$ and $B$ s.t. there exists and injection from $A$ to $B \implies |A| \leq |B|$. And if I know that $B$ is countable it follows that $A$ is countable, right? – Jan 08 '21 at 18:17
1

That's exactly right. – Andrew Kay Jan 09 '21 at 21:40

$f: [0,1] \rightarrow \mathbb{R}$ non-decreasing. Show $f \in L^1.$

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