Increasing Sets and Integral Limits
When measurable sets grow, their union records every point that eventually enters the sequence. For a nonnegative measurable function, integrating over those sets gives a corresponding increasing sequence of integrals. Continuity from below says that its limit is exactly the integral over the union. The statement allows infinite values and does not require the measure space to be finite or sigma-finite.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space, with no assumption of completeness, finiteness, or sigma-finiteness. Functions in the results below are measurable and take values in \([0,\infty]\). For \(E\in\mathcal{M}\), define the integral over \(E\) by \(\int_E f\,d\mu=\int f\chi_E\,d\mu\). Integrals and limits are understood in \([0,\infty]\), so equality may include the value \(+\infty\).
The key connection is between restricting a function to a set and taking a pointwise limit. If the sets increase, then \(f\chi_{E_n}\) increases pointwise to \(f\chi_E\). The Monotone Convergence Theorem, established earlier, applies to this sequence. It requires measurable, nonnegative functions increasing pointwise; it does not require the integrals to be finite.
Continuity from Below for Integrals
Proof. For every \(n\), let \(g_n=f\chi_{E_n}\) and let \(g=f\chi_E\). These functions are measurable and nonnegative, since \(f\) and the indicators of measurable sets are measurable. If \(x\in E_n\), then \(x\in E_{n+1}\), so \(g_n(x)=f(x)=g_{n+1}(x)\). If \(x\notin E_n\), then \(g_n(x)=0\le g_{n+1}(x)\). Thus \(g_n\le g_{n+1}\) everywhere. For each \(x\in E\), there is an index \(N\) with \(x\in E_N\); for all \(n\ge N\), \(g_n(x)=f(x)\). For \(x\notin E\), \(g_n(x)=0\) for every \(n\). Consequently \(g_n(x)\to g(x)\) at every \(x\). The hypotheses of the Monotone Convergence Theorem are satisfied, so \(\lim_n\int g_n\,d\mu=\int g\,d\mu\). By the definition of integration over a set, this is the asserted identity. \(\square\)
The result also explains how increasing functions fit into the picture. If measurable \(f_n:X\to[0,\infty]\) satisfy \(f_n\uparrow f\) pointwise, then the Monotone Convergence Theorem gives \(\lim_n\int f_n\,d\mu=\int f\,d\mu\). Continuity from below is its set-restriction form: take \(f_n=f\chi_{E_n}\). The distinction is useful: the sets need not have finite measure, and neither the restricted integrals nor the limiting integral need be finite.
Proof. Apply the theorem to the measurable function \(f=\chi_X\), which is nonnegative and equals \(1\) everywhere. Then \(\int_{E_n}\chi_X\,d\mu=\mu(E_n)\) and \(\int_E\chi_X\,d\mu=\mu(E)\). The theorem gives the claimed equality. \(\square\)
More generally, applying the theorem to \(f=\chi_A\), for any \(A\in\mathcal{M}\), gives \(\lim_n\mu(A\cap E_n)=\mu(A\cap E)\). This follows because \(\int_{E_n}\chi_A\,d\mu=\mu(A\cap E_n)\). Thus the same argument can track the measure of the part of an increasing set that lies inside a fixed measurable set.
Worked Applications
Worked Example: Indicators of an Increasing Union
On \(\mathbb{R}\) with Lebesgue measure \(m\), let \(E_n=[0,1-1/n]\) for integers \(n\ge2\). Find \(\lim_n\int\chi_{E_n}\,dm\) and identify the pointwise limit of the indicators.
The right endpoint \(1-1/n\) increases with \(n\), so \(E_n\subseteq E_{n+1}\). Every \(x\) with \(0\le x<1\) belongs to some \(E_n\): choose \(n\) large enough that \(1/n\le1-x\). No \(E_n\) contains \(1\), and none contains a point outside \([0,1]\). Therefore \(\bigcup_{n\ge2}E_n=[0,1)\), and \(\chi_{E_n}\) increases pointwise to \(\chi_{[0,1)}\).
Each interval \(E_n\) has Lebesgue measure \((1-1/n)-0=1-1/n\). Hence \(\int\chi_{E_n}\,dm=m(E_n)=1-1/n\), and this tends to \(1\). By continuity from below, \(\int\chi_{[0,1)}\,dm=m([0,1))=1\). Check: the direct interval length is \(1-0=1\), agreeing with the limit. The endpoint \(1\) is excluded from the union; its exclusion does not change the Lebesgue measure.
Worked Example: Integrating a Function over Expanding Sets
On \(\mathbb{R}\) with Lebesgue measure \(m\), define \(f=2\) on \([0,1]\), \(f=3\) on \((1,2]\), and \(f=0\) elsewhere. Let \(E_n=[0,2-1/n]\) for \(n\ge1\). Compute the limit of \(\int_{E_n}f\,dm\).
The sets increase, since \(2-1/n\le2-1/(n+1)\). Their union is \([0,2)\): every \(x\in[0,2)\) is eventually below \(2-1/n\), while \(2\) is never included. For \(n=1\), \(E_1=[0,1]\), so \(\int_{E_1}f\,dm=2\cdot1=2\). For \(n\ge2\), the part of \(E_n\) in \([0,1]\) has length \(1\), and the part in \((1,2-1/n]\) has length \((2-1/n)-1=1-1/n\). Therefore \[ \int_{E_n}f\,dm=2\cdot1+3\left(1-\frac1n\right)=5-\frac3n. \] As \(n\to\infty\), this tends to \(5\). Continuity from below gives \(\int_{[0,2)}f\,dm=5\). Check: integrating directly over \([0,2)\) gives \(2\cdot1+3\cdot1=5\). The omitted endpoint \(2\) has measure zero, and in this example \(f(2)=3\) is finite, so its omission contributes zero.
Worked Example: An Increasing Sequence of Functions
On \([0,1]\) with Lebesgue measure \(m\), set \(f_n(x)=1-(1-x)^n\) for \(n\ge1\). Verify that \(f_n\) increases, then compute the limit of its integrals.
For \(x\in[0,1]\), \(0\le1-x\le1\), so \((1-x)^{n+1}\le(1-x)^n\). Subtracting from \(1\) gives \(f_n(x)\le f_{n+1}(x)\), and \(f_n(x)\ge0\). If \(x>0\), then \(0\le1-x<1\), so \((1-x)^n\to0\) and \(f_n(x)\to1\). At \(x=0\), \(f_n(0)=0\) for every \(n\). Thus the pointwise limit is \(f=\chi_{(0,1]}\).
Direct integration gives \[ \int_0^1 f_n(x)\,dx =\int_0^1\bigl(1-(1-x)^n\bigr)\,dx =1-\frac{1}{n+1} =\frac{n}{n+1}. \] The limit is \(1\). The Monotone Convergence Theorem applies because every \(f_n\) is measurable, nonnegative, and the sequence increases pointwise; it follows that \(\int_0^1 f\,dm=1\). Check: \(f=\chi_{(0,1]}\), whose integral is the length of \((0,1]\), namely \(1\). The value at \(x=0\) does not alter the integral.
Worked Example: A Limit of Infinite Integrals
On \(\mathbb{R}\) with Lebesgue measure \(m\), take \(f(x)=1/x\) on \((0,1)\), and \(f(x)=0\) elsewhere. Let \(E_n=[1/n,1)\) for \(n\ge2\). Determine the limiting integral.
The sets increase and their union is \((0,1)\): each \(x\in(0,1)\) satisfies \(1/n\le x\) for all sufficiently large \(n\), while \(0\) is in no \(E_n\). For each \(n\ge2\), integration on the interval gives \[ \int_{E_n}f\,dm=\int_{1/n}^{1}\frac{1}{x}\,dx =\ln(1)-\ln(1/n)=\ln n. \] Since \(\ln n\to+\infty\), continuity from below gives \(\int_{(0,1)}f\,dm=+\infty\). Check: the improper integral \(\int_0^1 dx/x\) also diverges, as its integrals from \(1/n\) to \(1\) grow without bound. The theorem permits this infinite value; it does not assert that each integral is finite.
Why the Direction and Nonnegativity Matter
The theorem concerns increasing sets and nonnegative functions. Neither condition can be silently reversed. For a concrete failure of the decreasing-set analogue without a finite bound, take \((\mathbb{R},\mathcal{L},m)\), \(f=\chi_{\mathbb{R}}\), and \(E_n=[n,\infty)\). These sets decrease and have empty intersection. For every \(n\), \(\int_{E_n}f\,dm=m([n,\infty))=+\infty\), whereas \(\int_{\cap_nE_n}f\,dm=\int_{\varnothing}f\,dm=0\). Thus the limit of the integrals is not the integral over the intersection. This does not contradict continuity from below: the sets move in the opposite direction.
Likewise, nonnegativity is part of the stated result. For a signed function, expressions can involve cancellation, and the nonnegative Monotone Convergence Theorem cannot be applied just because the sets increase. Before using continuity from below, check that the sets are measurable and nested in the stated direction, and that the integrand is measurable and nonnegative. No finite-integral check is needed.
A common mistake is to infer that \(\int_{\cup_n E_n}f\,d\mu\) equals one of the finite-stage integrals, or that a finite limit follows automatically. For the expanding intervals \(E_n=[1/n,1)\) above, every finite-stage integral is \(\ln n\), while the union integral is \(+\infty\). The correct conclusion is that the integrals converge to the union integral in the extended nonnegative real numbers; that limit may be finite or infinite.
Check Your Understanding
For each question, identify the measure space and verify the hypotheses before applying continuity from below or Monotone Convergence.
- On \(\mathbb{R}\) with Lebesgue measure, let \(E_n=[0,2-1/n]\) for \(n\ge1\). Find their union and the limit of \(m(E_n)\).
- For a nonnegative measurable \(f\) on an arbitrary measure space and increasing measurable sets \(E_n\), explain why \(f\chi_{E_n}\) increases pointwise to \(f\chi_{\cup_nE_n}\).
- On \([0,1]\) with Lebesgue measure, let \(f_n(x)=x^n\) for \(n\ge1\). Does the sequence increase pointwise? If not, can the Monotone Convergence Theorem be applied in that order?
- Give a decreasing sequence of measurable sets whose measures do not converge to the measure of their intersection, and identify the missing direction hypothesis.
- If \(\mu(E_n)=n/(n+1)\) for an increasing sequence \(E_n\), what does continuity from below say about \(\mu(\cup_nE_n)\)?