Why Construct Simple Approximations?
In the definition of the Lebesgue integral of a nonnegative measurable function, simple functions provide the basic finite-valued building blocks. To use them effectively, it is helpful to have a concrete way to approximate an arbitrary function from below. The construction here rounds values down to dyadic levels and raises a truncation threshold as the index increases. Rounding down keeps each approximation below the function; nested dyadic grids and increasing truncation thresholds make the approximations increase.
Throughout, \((X,\mathcal{M},\mu)\) is any measure space. No assumption that \(\mu\) is finite, sigma-finite, or complete is needed. The function \(f:X\to[0,\infty]\) is measurable, and all convergence and inequalities in this tutorial are pointwise everywhere, not merely almost everywhere. A nonnegative simple function means a measurable function taking only finitely many values in \([0,\infty)\). The construction also applies when \(X\) is empty.
The cutoff at \(n\) ensures the output is finite even when \(f(x)=\infty\). The dyadic spacing \(2^{-n}\) gives a finite set of possible outputs: \(0,2^{-n},2\cdot2^{-n},\ldots,n\). Thus the formula is designed to produce simple functions, not just pointwise approximations.
Writing the Approximation as a Simple Function
The formula has a direct description by measurable level sets. For each fixed \(n\), the sets \(\{x:k2^{-n}\le f(x)<(k+1)2^{-n}\}\), for \(k=0,\ldots,n2^n-1\), together with \(\{x:f(x)\ge n\}\), are measurable and partition \(X\). The last set includes points where \(f(x)=\infty\). On the \(k\)-th set the approximation equals \(k2^{-n}\), and on the last set it equals \(n\). Therefore
Every coefficient and every level set in this finite sum is specified, including the zero level: the \(k=0\) term is zero and contributes nothing to the function. This representation proves directly that \(s_n\) is measurable and finite-valued, so it is a simple function. It also shows that no assumption about the measure of any level set is needed.
Worked Example: Approximating a Constant Value
Let \(X\) be any measurable space and let \(f(x)=5/6\) for every \(x\in X\). Compute the first few approximations. Since \(n\ge1>5/6\), the truncation does not alter this value.
For \(n=1\), \(2^1(5/6)=5/3\), whose floor is \(1\). Hence \(s_1=1/2\). For \(n=2\), \(2^2(5/6)=10/3\), whose floor is \(3\), so \(s_2=3/4\). For \(n=3\), \(2^3(5/6)=20/3\), whose floor is \(6\), so \(s_3=6/8=3/4\). For \(n=4\), \(2^4(5/6)=40/3\), whose floor is \(13\), so \(s_4=13/16\). For \(n=5\), \(2^5(5/6)=80/3\), whose floor is \(26\), so \(s_5=26/32=13/16\).
Thus the values are \(1/2,3/4,3/4,13/16,13/16\). Each is at most \(5/6\), and the sequence does not decrease. For example, the errors at \(n=2\) and \(n=4\) are \(5/6-3/4=1/12\) and \(5/6-13/16=1/48\), respectively. Both are smaller than the corresponding dyadic spacings \(1/4\) and \(1/16\). Check: the fourth value is \(13/16=39/48\), while \(5/6=40/48\), so the error is indeed \(1/48\).
The Increasing Approximation Theorem
Proof. The finite level-set representation shows that every \(s_n\) is measurable and takes only finitely many finite values. Since a floor is nonnegative for a nonnegative argument, \(s_n\ge0\). Also, \(\lfloor r\rfloor\le r\) for every finite \(r\), so \(s_n(x)\le\min(f(x),n)\le f(x)\), with the same inequalities valid when \(f(x)=\infty\).
To prove that the sequence increases, fix \(x\) and write \(t=f(x)\). The function \(u\mapsto\lfloor 2^n u\rfloor/2^n\) is nondecreasing in \(u\). Moreover, for every finite \(u\ge0\), the dyadic grids are nested, and $$ \frac{\lfloor 2^n u\rfloor}{2^n} \le \frac{\lfloor 2^{n+1}u\rfloor}{2^{n+1}}. $$ Indeed, if \(j=\lfloor2^n u\rfloor\), then \(2^{n+1}u\ge2j\), so \(\lfloor2^{n+1}u\rfloor\ge2j\). Now \(\min(t,n)\le\min(t,n+1)\). First refine the grid at \(\min(t,n)\), then increase the truncated argument; these two inequalities give \(T_n(t)\le T_{n+1}(t)\). This reasoning includes \(t=\infty\), since the truncated arguments are then \(n\) and \(n+1\).
It remains to verify the limit. If \(t<\infty\), choose \(n>t\). Then \(\min(t,n)=t\), and the floor inequality gives $$ 0\le t-T_n(t)<2^{-n}. $$ The right side tends to zero, so \(T_n(t)\to t\). If \(t=\infty\), then \(T_n(t)=2^{-n}\lfloor2^n n\rfloor=n\), which tends to infinity. Applying these conclusions with \(t=f(x)\) proves the claimed pointwise limit for every \(x\). \(\square\)
Worked Example: The Function f(x) = x on [0, 2]
Take \(X=[0,2]\), its Lebesgue sigma-algebra, and Lebesgue measure \(m\), and let \(f(x)=x\). The level-set formula gives a simple function on each dyadic interval below the cutoff \(n\), with value \(k/2^n\) on \([k/2^n,(k+1)/2^n)\cap[0,2]\); values at the cutoff are assigned \(n\).
At \(x=5/8\), for \(n=1\), the scaled value is \(2(5/8)=5/4\), with floor \(1\), giving \(s_1(5/8)=1/2\). For \(n=2\), the scaled value is \(4(5/8)=5/2\), with floor \(2\), giving \(s_2(5/8)=2/4=1/2\). For \(n=3\), the scaled value is \(8(5/8)=5\), giving \(s_3(5/8)=5/8\). At \(x=3/2\), the first cutoff gives \(s_1(3/2)=1\), because \(3/2\ge1\). At \(n=2\), \(4(3/2)=6\), so \(s_2(3/2)=6/4=3/2\). At \(n=3\), \(8(3/2)=12\), so \(s_3(3/2)=12/8=3/2\). Finally, at \(x=2\), the first two cutoffs give \(s_1(2)=1\) and \(s_2(2)=2\), and for \(n\ge2\) the value is exactly \(2\).
Check: at \(5/8\), the displayed approximations are \(1/2,1/2,5/8\), all at most \(5/8\). At \(3/2\), they are \(1,3/2,3/2\), all at most \(3/2\). At \(2\), they are \(1,2,2\), all at most \(2\). These pointwise checks agree with the theorem and illustrate that some values can become exact before the approximation index reaches the value.
Worked Example: A Measurable Function with an Infinite Value
Let \(A,B,C\in\mathcal{M}\) be pairwise disjoint sets whose union is \(X\), and define \(f\) to equal \(3/8\) on \(A\), \(3/2\) on \(B\), and \(\infty\) on \(C\). This is an extended-real-valued measurable function because its level sets are measurable.
For \(n=1\), the value on \(A\) rounds down to zero: \(2(3/8)=3/4\), whose floor is \(0\). On \(B\), truncation replaces \(3/2\) by \(1\), and \(2(1)=2\), giving \(s_1=1\). On \(C\), the value \(\infty\) is also truncated to \(1\), giving \(s_1=1\). Thus \(s_1=\chi_B+\chi_C\).
For \(n=2\), on \(A\) the scaled value is \(4(3/8)=3/2\), so the rounded value is \(1/4\). On \(B\), \(4(3/2)=6\), giving \(6/4=3/2\). On \(C\), truncation gives \(2\), which is exactly on the grid, so the approximation is \(2\). Therefore \(s_2=\frac14\chi_A+\frac32\chi_B+2\chi_C\).
For \(n=3\), on \(A\), \(8(3/8)=3\), giving \(3/8\); on \(B\), \(8(3/2)=12\), giving \(12/8=3/2\); on \(C\), truncation gives \(3\), so \(s_3=\frac38\chi_A+\frac32\chi_B+3\chi_C\). Check: the values on \(A\) are \(0,1/4,3/8\); on \(B\) they are \(1,3/2,3/2\); and on \(C\) they are \(1,2,3\). Each list is nondecreasing and bounded above by the corresponding value of \(f\), with the values on \(C\) tending to infinity.
Why Truncation and Nested Grids Matter
A tempting alternative is to set \(r_n(x)=2^{-n}\lfloor2^n f(x)\rfloor\) without truncation. This can fail in two distinct ways. If \(f(x)=\infty\), the floor in that formula is not defined as a finite integer. Even when \(f\) is finite everywhere, \(r_n\) need not be simple: for \(f(x)=1/x\) on \((0,1]\), the values of \(2^{-n}\lfloor2^n/x\rfloor\) are unbounded as \(x\) approaches zero, so the function takes infinitely many values for every fixed \(n\). A finite-valued simple approximation requires a cutoff as well as rounding.
Another potential error is to round to a grid that does not refine the preceding grid, or to use unrelated truncation levels. The proof of monotonicity depends on both structural choices: every multiple of \(2^{-n}\) is also a multiple of \(2^{-(n+1)}\), and \(\min(t,n)\le\min(t,n+1)\). Neither the pointwise convergence alone nor the fact that each function is simple would guarantee an increasing sequence.
The construction is useful because it provides explicit finite-valued approximations from below for every nonnegative measurable function, including unbounded functions and functions taking the value infinity. It does not itself assert a limit formula for integrals; such a conclusion requires a separate convergence theorem and its hypotheses. Here the conclusion is exactly the pointwise construction and its verified approximation properties.
Check Your Understanding
Use the definition of \(T_n\) and keep track of whether truncation is active at each point.
- For a constant function \(f=7/8\), compute \(s_1,s_2,s_3\), and verify that each is at most \(7/8\).
- If \(f(x)=\infty\), compute \(s_n(x)\) for every \(n\). Does the sequence converge to a finite or infinite value?
- Explain why \(s_n\) has only finitely many values even if \(f\) is unbounded.
- For a finite value \(t\), state the error bound that applies once \(n>t\).
- Give one reason that rounding down without truncation may fail to give a simple function.