A sequence of simple functions converges uniformly to $f$

Question

The following is a problem from Carothers' Real Analysis:

Suppose $f$ a nonnegative, bounded, (Lebesgue) measurable function on $[a,b]$, with $0\leq f\leq M$. Let $$E_{n,k}=\left\{\frac{kM}{2^n}\leq f < \frac{(k+1)M}{2^n}\right\}$$ for each $n\in\mathbb{N}$ and $k=0,1,2,\ldots, 2^n$ and set $$\varphi_n=\sum_{k=0}^{2^n}\frac{kM}{2^n}\chi_{E_{n,k}}$$

Prove that $0\leq \varphi_n\leq\varphi_{n+1}\leq f$ and that $0\leq f-\varphi_n\leq 2^{-n}M$ for each $n$. Then it is possible to conclude that the sequence of simple functions $(\varphi_n)\to f$ uniformly on $[a,b]$.

(where $\chi_{E_{n,k}}$ is the characteristic/indicator function of $E_{n,k}$)

Thoughts:

I'm not really sure how to start, so I'll just list some information that could be of help.

• $f$ Lebesgue measurable means that $f^{-1}((\alpha,\infty))$ is measurable as well for all $\alpha\in\mathbb{R}$

• $\varphi_n\geq 0$ because the characteristic function only has an image of $0$ or $1$, and $\frac{kM}{2^n}$ is non-negative, so the sum of their product must also be nonnegative

• I can split up $E_{n,k}$ into the union of two disjoint sets, $E_{n,k}=E_{n+1,2k}\cup E_{n+1,2k+1}$ (This is a hint given by Carothers under the problem.)

• $$\varphi_n=\sum_{k=0}^{2^n}\frac{kM}{2^n}\chi_{E_{n,k}}=\frac{M}{2^n}\sum_{k=0}^{2^n}k\chi_{E_{n,k}}$$ so it makes sense why in the second inequality, $0$ is the lower bound and $2^{-n}\cdot M$ is the upper bound.

Any help to start would be appreciated. Thanks.

Answer 1

For the first part, it suffices to prove that $\varphi_n \leq \varphi_{n+1}$ on $E_{n,k}$ for all $k = 1, \dots, 2^n$ and any fixed $n$. (why?)

Here's where to use $E_{k,n} = E_{2k,n+1}\cup E_{2k+1,n+1}.$ Simply draw the functions $\varphi_n$ and $\varphi_{n+1}$ to see what's going on. \begin{align*} \varphi_n\chi_{E_{k,n}} &= \frac{kM}{2^n}\chi_{E_{k,n}}\\ &= \frac{kM}{2^n}\chi_{E_{2k,n+1}}+\frac{kM}{2^n}\chi_{E_{2k+1,n+1}}\\ &\leq \frac{2kM}{2^{n+1}}\chi_{E_{2k,n+1}}+\frac{(2k+1)M}{2^{n+1}}\chi_{E_{2k+1,n+1}}\\ &= \varphi_{n+1} \chi_{E_{k,n}}. \end{align*} This observation should help you with $0 \leq f - \varphi_n\leq 2^{-n}M$.

For uniform convergence, let $\epsilon > 0$ and choose $N$ so large that $2^{-N}M \leq \epsilon$. Note that by the previous assertion and $f \geq \varphi_n$, we have $$|f-\varphi_n| \leq 2^{-N}M \leq \epsilon.$$