Approximating Integral Values from Below

Lower Bounds Built from Simple Functions

A simple function replaces a possibly complicated measurable function by finitely many values on measurable sets. When those values never exceed the function, its integral is a certified lower bound. This gives both a conceptual description of the nonnegative Lebesgue integral and a practical method: choose the sets and values explicitly, compute the simple-function integral, and estimate what remains below the target.

Throughout the general discussion, $(X,\mathcal{M},\mu)$ is a measure space, with no assumption that $\mu$ is finite or sigma-finite. Functions are measurable and take values in $[0,\infty]$. A nonnegative simple function is a measurable function taking only finitely many finite values; it can be written as $\varphi=\sum_{j=1}^r a_j\chi_{E_j}$, where the $a_j\ge0$ and the measurable sets $E_j$ are disjoint. Its integral is $\int\varphi\,d\mu=\sum_{j=1}^r a_j\mu(E_j)$, with infinite values allowed. We use the definition of the nonnegative integral as the supremum over simple minorants:

\int f\,d\mu = \sup\left\{\int\varphi\,d\mu: \varphi\text{ is nonnegative simple and }0\le\varphi\le f\right\}.

The inequality $\varphi\le f$ is pointwise here. The displayed supremum can be finite or infinite. In particular, a lower-bound construction does not require that the supremum be attained by one simple function.

What the Supremum Guarantees

Proposition: For any measure space $(X,\mathcal{M},\mu)$ and measurable $f:X\to[0,\infty]$, write $I=\int f\,d\mu$. If $I<\infty$, then for every $\varepsilon>0$ there is a nonnegative simple $\varphi\le f$ such that $I-\varepsilon<\int\varphi\,d\mu\le I$. If $I=\infty$, then for every finite $C\ge0$ there is a nonnegative simple $\varphi\le f$ with $\int\varphi\,d\mu>C$.

Proof. First suppose $I<\infty$. The set of integrals in the defining supremum is nonempty because the zero function is a simple minorant. If no such $\varphi$ had integral greater than $I-\varepsilon$, then $I-\varepsilon$ would be an upper bound for all those integrals, contradicting that their supremum is $I$. Every candidate integral is at most $I$, again by the definition of supremum. This proves the finite case, including $I=0$: for any $\varepsilon>0$, the zero minorant has integral $0>I-\varepsilon$. If $I=\infty$, the set of candidate integrals is unbounded above; otherwise its finite upper bound would contradict its infinite supremum. Thus some candidate has integral greater than $C$. $\square$

This proposition guarantees that lower bounds can be made arbitrarily accurate when the integral is finite, but does not itself prescribe the minorant. For a bounded function on a finite-measure space, rounding down supplies an explicit choice and an error estimate.

Theorem (Finite-grid lower approximation): Let $(X,\mathcal{M},\mu)$ be a measure space with $\mu(X)<\infty$. Suppose $f:X\to[0,B]$ is measurable and real-valued, where $B<\infty$. For $\delta>0$, define $\varphi_\delta(x)=\delta\lfloor f(x)/\delta\rfloor$. Then $\varphi_\delta$ is a nonnegative simple function, $0\le\varphi_\delta\le f$, and $$ 0\le \int f\,d\mu-\int\varphi_\delta\,d\mu\le\delta\,\mu(X). $$

Proof. Because $0\le f\le B$, the integer $\lfloor f(x)/\delta\rfloor$ takes only finitely many values, all nonnegative. For each such integer $k$, the set where it equals $k$ is measurable: it is $\{k\delta\le f<(k+1)\delta\}$, with the final set adjusted to include any values equal to $B$. Thus $\varphi_\delta$ is simple. The defining property of the floor gives $0\le f(x)-\varphi_\delta(x)<\delta$ at every $x$. In particular, $0\le\varphi_\delta\le f\le\varphi_\delta+\delta\chi_X$. Monotonicity and additivity of the nonnegative integral, and the formula for the integral of a constant, now give $$ \int\varphi_\delta\,d\mu \le\int f\,d\mu \le\int\varphi_\delta\,d\mu+\delta\mu(X). $$ All terms are finite because $f\le B$ and $\mu(X)<\infty$. Subtracting $\int\varphi_\delta\,d\mu$ proves the estimate. $\square$

The bound is useful even if it is not sharp: choosing $\delta<\varepsilon/\mu(X)$ makes the error less than $\varepsilon$ when $0<\mu(X)<\infty$. If $\mu(X)=0$, both integrals are zero. This theorem is a one-function rounding estimate; it does not require an interchange of a limit and an integral.

Worked Example: A Dyadic Minorant for f(x) = x

On $([0,1],\mathcal{L},m)$, let $N=2^n$ for a positive integer $n$. Define $\varphi_N(x)=k/N$ on $[k/N,(k+1)/N)$, for $k=0,\ldots,N-1$, and set $\varphi_N(1)=1$. This is a nonnegative simple function. On each half-open interval, $k/N\le x$, and at $x=1$ the two values agree, so $\varphi_N\le f$ everywhere.

Each interval has Lebesgue measure $1/N$, while the singleton $\{1\}$ has measure zero. Therefore the exact lower bound is

\int_{[0,1]}\varphi_N\,dm =\sum_{k=0}^{N-1}\frac{k}{N}\frac{1}{N} =\frac{1}{N^2}\frac{(N-1)N}{2} =\frac{N-1}{2N}.

The integral of $f(x)=x$ on $[0,1]$, computed directly, is $1/2$. Hence the error is $1/2-(N-1)/(2N)=1/(2N)$. Check: for $N=4$, the four interval contributions are $0$, $1/16$, $2/16$, and $3/16$, so the lower integral is $6/16=3/8$; the gap from $1/2$ is $1/8=1/(2\cdot4)$, as the formula predicts.

Worked Example: A Lower Sum for a Decreasing Function

On $([0,1],\mathcal{L},m)$, take $f(x)=1/(1+x)$. Divide $[0,1]$ into four intervals of length $1/4$, and on each interval use the value of $f$ at its right endpoint. Since $f$ is decreasing, these values do not exceed $f(x)$ on their respective intervals. Endpoint assignments do not affect the integral because a finite set has Lebesgue measure zero.

The lower simple function has values $4/5,\,2/3,\,4/7,\,1/2$, in that order. Each interval has measure $1/4$, so its integral is

\frac14\left(\frac45+\frac23+\frac47+\frac12\right) =\frac14\left(\frac{168+140+120+105}{210}\right) =\frac{533}{840}.

Direct integration gives $\int_0^1(1+x)^{-1}\,dx=\ln2$. For an error check, use the left endpoint values on the same partition to obtain an upper simple bound with integral $$ \frac14\left(1+\frac45+\frac23+\frac47\right) =\frac14\left(\frac{420+336+280+240}{420}\right) =\frac{319}{420}. $$ The difference between this upper sum and the lower sum is $319/420-533/840=105/840=1/8$. Thus the integral lies between the two sums, and the lower bound’s error is at most $1/8$. Check: both sums use the same four intervals; the right-endpoint values are below the left-endpoint values, and their integral difference is exactly $1/8$.

Separating Rounding from Truncation

On a domain of infinite measure, a uniform rounding error over the whole domain need not have a finite integral. A useful construction can instead restrict attention to a finite interval, use a simple minorant there, and account separately for the omitted tail. The next example computes both contributions exactly.

Worked Example: Truncating and Rounding e to the Negative x

On $([0,\infty),\mathcal{L},m)$, let $f(x)=e^{-x}$. Use three intervals of length $\ln2$, and set $\varphi(x)=2^{-(k+1)}$ on $[k\ln2,(k+1)\ln2)$ for $k=0,1,2$, with $\varphi(x)=0$ for $x\ge3\ln2$. On each interval $e^{-x}\ge e^{-(k+1)\ln2}=2^{-(k+1)}$, and beyond the intervals $0\le e^{-x}$. Thus $\varphi\le f$ everywhere.

The three intervals each have measure $\ln2$, so the minorant’s integral is $$ \int\varphi\,dm =(\ln2)\left(\frac12+\frac14+\frac18\right) =\frac{7\ln2}{8}. $$ The target integral is $\int_0^\infty e^{-x}\,dx=1$, evaluated from the antiderivative $-e^{-x}$. The omitted tail is $\int_{3\ln2}^{\infty}e^{-x}\,dx=e^{-3\ln2}=1/8$. On the retained interval $[0,3\ln2]$, the exact gap is $7/8-7\ln2/8=7(1-\ln2)/8$. Adding the tail gives the total error $1-7\ln2/8$. Check: the retained integral is $1-1/8=7/8$; subtracting the lower value $7\ln2/8$ and then adding the tail $1/8$ yields $1-7\ln2/8$, as required.

A Common Construction Mistake

To obtain a lower bound, values must be rounded down, not up. For instance, on $([0,1],\mathcal{L},m)$, consider $f(x)=x$ and the partition into four intervals. Rounding each interval’s left endpoint value up to the next multiple of $1/4$ gives the value $1/4$ on $[0,1/4)$. At $x=0$, this proposed function has value $1/4$, which is greater than $f(0)=0$. It is therefore not a minorant, and its integral cannot be used as a certified lower bound by the supremum definition.

The correction is to use the left endpoint value itself, or to round $f(x)$ downward to the grid. In the four-interval construction for $f(x)=x$, the valid values are $0,1/4,1/2,3/4$, giving integral $3/8$, not a value obtained by rounding up. More generally, before integrating a candidate simple function, check the pointwise inequality on every set in its defining partition, including endpoints where the definition assigns a value.

These constructions matter because they turn the supremum definition into usable estimates. On a finite-measure space, the finite-grid theorem supplies a controlled error for bounded functions. For a specific function, a partition adapted to its behavior can give a sharper bound, while a separate tail estimate handles an unbounded domain. In each case, the logical order is the same: verify the pointwise minorant, compute its integral from the measures of its level sets, and then bound the difference from the target.

Check Your Understanding

For each question, identify the measure space and verify the minorant condition before calculating its integral.

On $[0,1]$ with Lebesgue measure, divide into $N$ equal intervals and use the left endpoint values for $f(x)=x$. Compute the lower integral and its exact error.
On a finite-measure space with $0\le f\le B$, explain why rounding $f$ down in increments of $\delta$ gives a simple function and an integral error at most $\delta\mu(X)$.
For $f(x)=1/(1+x)$ on $[0,1]$, construct lower and upper sums using $N$ equal intervals. Why does monotonicity order the two sums?
For $f(x)=e^{-x}$ on $[0,\infty)$, use $N$ intervals of length $\ln2$, with right-endpoint values, and set the minorant to zero afterward. Compute its integral and the omitted tail.
Give a point where rounding $f(x)=x$ upward to multiples of $1/4$ on $[0,1/4)$ violates the minorant condition.

Approximating Integral Values from Below

What You'll Learn