Proof that an $n$th degree Lebesgue measure of a function is less than or equal to $(n+1)$th degree Lebesgue measure of the function

Question

We know that the Lebesgue measure of a nonnegative function is the limit (as $n \to \infty$) of the quantity $$S_nf = \sum_{j=0}^\infty \frac{j}{2^n}m(\{x: \frac{j}{2^n} \le f(x) < \frac{j+1}{2^n}\})$$ The question asks: "Prove that, for a function $f: \Bbb R \to \Bbb R^{\ge 0}$ we have $S_nf \le S_{n+1}f$."

I am supposed to use the fact that: given a function $f: A \to B$ and $C, D \subset B$, where $C$ and $D$ are disjoint, then the preimage of $C$ and $D$ also are disjoint. We also are allowed to use some basic properties of Lesbesgue measure of a set, such as $m(A \cup B) \le m(A) + m(B)$.

I get the intuition that the Lebesgue measure is similar to the "lower sum" of a function, but I do not know how to answer the question rigorously. I tried setting $j = 2u$, but to no avail.

Answer 1

Hint: Using the notation $$ A(n,j):=\left\{x:\frac j{2^n}\le f(x)<\frac{j+1}{2^n}\right\}, $$ the key step is to show that $$A(n,j)=A(n+1, 2j)\cup A(n+1, 2j+1), $$ a disjoint union. So each term of the summation that defines $S_n f$ yields two terms. Follow the provided advice. After some bookkeeping you should obtain the upper bound $S_{n+1} f$.