How to Approximate Simple Functions by Step Functions Almost Uniformly

A measurable simple function takes only finitely many values. On a finite interval, you can approximate it by step functions that converge almost uniformly: after removing a set of arbitrarily small measure, the approximations converge uniformly on what remains.

The key is to approximate each level set by a finite union of intervals, with errors small enough that their measures add to a finite total. The summability of those errors controls all later approximations at once.

The setting and the result

Let $I=[a,b]$ be a finite interval, equipped with Lebesgue measure, and let $f:I\to\mathbb{R}$ be a measurable simple function. Write its distinct values and level sets as

$$f=\sum_{j=1}^{r} c_j\mathbf{1}_{E_j},$$

where the $c_j$ are constants and the $E_j$ are measurable and pairwise disjoint. The level set where $f=0$ may be included or omitted.

A step function here means a function constant on each interval of some finite partition of $I$. Equivalently, it can be written as a finite linear combination of indicators of finite unions of intervals. The claim is that there is a sequence of such step functions $s_n$ that converges to $f$ almost uniformly.

Almost-uniform convergence means that for every $\delta>0$, there is a measurable set $A\subseteq I$ with measure less than $\delta$ such that $s_n\to f$ uniformly on $I\setminus A$.

Construct the approximations

Every Lebesgue-measurable subset $E$ of a finite interval can be approximated in measure by a finite union of intervals. In particular, for any $\eta>0$, there is such a union $G$ with

$$m(E\mathbin{\triangle}G)<\eta,$$

where $E\mathbin{\triangle}G$ is the symmetric difference: the points belonging to exactly one of the two sets. This is a standard consequence of the regularity of Lebesgue measure.

For each $n$ and each level set $E_j$, choose a finite union of intervals $G_{j,n}$ such that

$$m(E_j\mathbin{\triangle}G_{j,n})<\frac{2^{-n}}{r}.$$

Now define

$$s_n=\sum_{j=1}^{r}c_j\mathbf{1}_{G_{j,n}}.$$

Each $s_n$ is a step function: there are only finitely many interval endpoints among the sets $G_{j,n}$, and those endpoints partition $I$ into finitely many subintervals on which every indicator, and hence $s_n$, is constant.

Let

$$B_n=\bigcup_{j=1}^{r}(E_j\mathbin{\triangle}G_{j,n}).$$

By subadditivity of measure,

$$m(B_n)\leq\sum_{j=1}^{r}m(E_j\mathbin{\triangle}G_{j,n})<2^{-n}.$$

At every point outside $B_n$, all the relevant indicators agree, so $s_n(x)=f(x)$. Thus any point that belongs to only finitely many $B_n$ satisfies $s_n(x)=f(x)$ for all sufficiently large $n$.

Why the convergence is almost uniform

Fix $\delta>0$. Choose $N$ large enough that

$$\sum_{n=N}^{\infty}2^{-n}<\delta,$$

and remove the set $A=\bigcup_{n=N}^{\infty}B_n$. Its measure satisfies

$$m(A)\leq\sum_{n=N}^{\infty}m(B_n)\leq\sum_{n=N}^{\infty}2^{-n}<\delta.$$

For every $x\in I\setminus A$ and every $n\geq N$, we have $x\notin B_n$, so $s_n(x)=f(x)$. In fact, the functions agree everywhere on $I\setminus A$ for every $n\geq N$. Therefore

$$\sup_{x\in I\setminus A}|s_n(x)-f(x)|=0\qquad(n\geq N),$$

which proves uniform convergence there. Since $\delta$ was arbitrary, the convergence is almost uniform.

Worked example

Suppose $I=[0,1]$ and

$$f=3\mathbf{1}_E-2\mathbf{1}_F,$$

where $E$ and $F$ are disjoint measurable sets. For each $n$, choose finite unions of intervals $G_n$ and $H_n$ satisfying

$$m(E\mathbin{\triangle}G_n)<2^{-n-1},\qquad m(F\mathbin{\triangle}H_n)<2^{-n-1}.$$

Then $s_n=3\mathbf{1}_{G_n}-2\mathbf{1}_{H_n}$ is a step function. The set on which either indicator might be wrong has measure less than $2^{-n}$. The union of these error sets over $n\geq N$ has measure at most $\sum_{n=N}^{\infty}2^{-n}=2^{1-N}$. Given any $\delta>0$, choose $N$ so that $2^{1-N}<\delta$. Outside that union, $s_n=f$ for every $n\geq N$, giving uniform convergence on a set whose complement has measure less than $\delta$.

The approximating interval unions $G_n$ and $H_n$ need not be disjoint. That does not affect the argument: outside the error sets, their indicators match the original level-set indicators, which are what determine $f$.

Common mistakes

• Using only that the errors tend to zero. The condition $m(B_n)\to0$ alone does not make the union of all later error sets small. Choose errors with a finite sum, as above.
• Confusing almost-uniform with pointwise convergence almost everywhere. Almost-uniform convergence requires uniform convergence after removing one small set. The tail-union argument supplies that stronger control.
• Ignoring the finite-measure setting. The interval approximation and the stated almost-uniform conclusion are being used on a finite-measure domain. On an infinite-measure domain, first specify an appropriate finite-measure setting or formulate a local version.
• Assuming this applies to every measurable function. The argument uses the fact that a simple function has finitely many level sets. Approximation results for general measurable functions require additional steps.

The essential mechanism is summability: approximate each level set with errors whose total measure is finite, then discard the union of all errors after a sufficiently late index.