A Unit Interval Can Move Away Without Shrinking
Pointwise convergence tests one fixed point at a time. An integral, by contrast, accounts for the values of a function across the whole space. These viewpoints can disagree when the region where a function is nonzero moves farther and farther away: every fixed point is eventually missed, even though the region keeps the same measure.
Throughout, the measure space is \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This measure space is complete and sigma-finite, but not finite. All functions below are real-valued, measurable, and nonnegative. Unless stated otherwise, convergence of functions means pointwise convergence at every \(x\in\mathbb{R}\), not merely convergence almost everywhere. For a measurable set \(E\), its indicator is \(\chi_E(x)=1\) when \(x\in E\) and \(0\) otherwise; if \(m(E)<\infty\), then \(\int\chi_E\,dm=m(E)\).
Worked Example: Unit Intervals Moving to the Right
For \(n\ge1\), let \(I_n=[n,n+1)\) and \(f_n=\chi_{I_n}\). The left endpoint is included and the right endpoint is excluded. Thus \(f_n(x)=1\) exactly when \(n\le x<n+1\).
Fix any \(x\in\mathbb{R}\). Choose an integer \(N\ge1\) with \(N>x\). For every \(n\ge N\), we have \(n\ge N>x\), so \(x<n\) and therefore \(x\notin[n,n+1)\). Consequently \(f_n(x)=0\) for all \(n\ge N\), and \(f_n(x)\to0\). This holds for every real \(x\), so the pointwise limit is the zero function.
For each \(n\), the interval length is \((n+1)-n=1\), and hence \(m(I_n)=1\). The indicator integral formula gives
On the other hand, \(\int_{\mathbb{R}}0\,dm=0\). Thus \(\lim_{n\to\infty}\int f_n\,dm=1\), while the integral of the pointwise limit is \(0\). Check: for \(n=1,2,3\), the intervals are \([1,2),[2,3),[3,4)\); each has length \(2-1=3-2=4-3=1\), as the formula predicts.
Why This Sequence Is Not Monotone
The Monotone Convergence Theorem applies to a sequence of nonnegative measurable functions that is increasing pointwise. The functions in the example are nonnegative and measurable, but they are not increasing. Indeed, at \(x=3/2\), \(f_1(3/2)=1\) and \(f_2(3/2)=0\), so \(f_1\le f_2\) fails. They are not decreasing either: at \(x=5/2\), \(f_1(5/2)=0\) and \(f_2(5/2)=1\), so \(f_2\le f_1\) fails.
The set supports move rather than nest. In particular, \([1,2)\) and \([2,3)\) are disjoint, but neither the pointwise limit nor the integrals are obtained by assuming that the supports become smaller. Each individual interval still has measure one. Pointwise convergence says that a fixed point eventually lies outside the supports; it does not say that the supports have small measure.
Proof. Each interval \([a_n,a_n+1)\) is Lebesgue measurable and has measure \((a_n+1)-a_n=1\), so the indicator integral formula gives \(\int g_n\,dm=1\). Now fix \(x\in\mathbb{R}\). Since \(a_n\to+\infty\), there is an \(N\) such that \(a_n>x\) for all \(n\ge N\). For each such \(n\), \(x<a_n\), so \(x\notin[a_n,a_n+1)\) and \(g_n(x)=0\). Therefore \(g_n(x)\to0\). This proves pointwise convergence at every \(x\). Finally, because \(a_n\to+\infty\), there exists \(n\) with \(a_n>a_1+1\). The point \(x=a_1+\tfrac12\) belongs to \([a_1,a_1+1)\) but not to \([a_n,a_n+1)\), so \(g_1(x)=1>0=g_n(x)\). Thus the sequence is not increasing. \(\square\)
Worked Example: Intervals Moving Faster to the Right
Set \(J_n=[2n,2n+1)\) and \(h_n=\chi_{J_n}\), for \(n\ge1\). The endpoints are treated with the same convention: \(h_n(x)=1\) precisely for \(2n\le x<2n+1\). Given a fixed \(x\), choose \(N\ge1\) with \(2N>x\). For \(n\ge N\), \(2n\ge2N>x\), so \(x\notin J_n\). Hence \(h_n(x)\to0\) at every real \(x\).
The interval length is \((2n+1)-2n=1\); therefore \(\int h_n\,dm=m(J_n)=1\) for every \(n\), whereas the integral of the limit is zero. Check: at \(n=2\), \(J_2=[4,5)\), whose length is \(5-4=1\), so \(\int h_2\,dm=1\). This is an instance of the proposition with \(a_n=2n\), which tends to \(+\infty\).
The Limit Can Be Nonzero as Well
The discrepancy is not restricted to sequences whose pointwise limit is zero. Keep a fixed interval in every set and add a second unit interval that escapes to the right. The escaping part disappears at each fixed point, but its integral remains one.
Worked Example: A Fixed Interval Plus an Escaping Interval
For \(n\ge1\), define \(E_n=[0,1)\cup[n+1,n+2)\) and \(q_n=\chi_{E_n}\). The two intervals are disjoint: the first ends at \(1\), excluded, while the second begins at \(n+1\ge2\). Both intervals have length one. Thus \(m(E_n)=1+1=2\) and \(\int q_n\,dm=2\) for every \(n\).
Fix \(x\in\mathbb{R}\). If \(x\in[0,1)\), then \(x\in E_n\) for every \(n\), because the fixed interval is included each time. If \(x\notin[0,1)\), choose an integer \(N\ge1\) with \(N+1>x\). For \(n\ge N\), the left endpoint \(n+1\) of the moving interval is greater than \(x\), so \(x\notin[n+1,n+2)\); by assumption \(x\) is also outside \([0,1)\). Thus \(q_n(x)=0\) for all \(n\ge N\). These two cases show that \(q_n\to\chi_{[0,1)}\) pointwise everywhere.
The pointwise limit has integral \(m([0,1))=1\), while the integrals of the sequence are all \(2\). Check: when \(n=1\), \(E_1=[0,1)\cup[2,3)\); the disjoint interval lengths are \(1-0=1\) and \(3-2=1\), giving \(m(E_1)=2\). The limit retains only \([0,1)\), of measure \(1-0=1\).
Pointwise, Uniform, and In-Measure Convergence
For the first sequence \(f_n=\chi_{[n,n+1)}\), pointwise convergence to zero does not imply uniform convergence. Uniform convergence to zero would require that for every \(\varepsilon>0\), there is an \(N\) such that \(|f_n(x)|<\varepsilon\) for every \(x\in\mathbb{R}\) and every \(n\ge N\). Taking \(\varepsilon=\tfrac12\), for every \(n\) the point \(x=n+\tfrac12\) satisfies \(f_n(x)=1\), so this condition fails for every proposed \(N\).
It does not converge to zero in measure on \((\mathbb{R},\mathcal{L},m)\), either. By definition, convergence in measure to zero would require \(m(\{x:|f_n(x)|>\varepsilon\})\to0\) for each \(\varepsilon>0\). At the fixed threshold \(\varepsilon=\tfrac12\), the strict inequality \(f_n(x)>\tfrac12\) holds exactly on \([n,n+1)\), so
for every \(n\). The tail measures therefore do not tend to zero. Likewise, for any \(1\le p<\infty\), \(\|f_n\|_p=(\int |f_n|^p\,dm)^{1/p}=(1)^{1/p}=1\), not zero. These computations distinguish pointwise convergence from other modes of convergence; they are not alternative hypotheses that make the integrals converge in this example.
Fatou's lemma, established earlier in the course, is consistent with the calculation. Its hypothesis here is satisfied: every \(f_n\) is nonnegative and measurable on \((\mathbb{R},\mathcal{L},m)\). It gives \(\int\liminf_n f_n\,dm\le\liminf_n\int f_n\,dm\), which in this case reads \(0\le1\). This lower bound does not assert equality and does not permit passing the pointwise limit through the integral.
Common Mistake: Confusing Location with Measure
A tempting but incorrect calculation is: “\(f_n(x)\to0\) for each \(x\), so \(\int f_n\,dm\to\int0\,dm=0\).” It treats pointwise convergence as if it justified interchanging the limit and the integral. The explicit computation shows the error: for every \(n\), the support is an interval of length \((n+1)-n=1\), and therefore \(\int f_n\,dm=1\). The limit function is zero and has integral zero, so the two quantities differ.
The correction is to check the hypotheses of a convergence theorem before using it. Here the functions are nonnegative and measurable, but they are not increasing, so the Monotone Convergence Theorem does not apply. The failure is not a defect in the integral formula: each integral is computed exactly. It is the unsupported passage from pointwise convergence to convergence of integrals that is wrong.
The useful conclusion is precise: pointwise convergence alone does not force convergence of integrals on an infinite-measure space. In these examples, a fixed point eventually misses the moving interval, but the interval never loses measure. To conclude that integrals converge, one needs additional hypotheses supplied by an applicable convergence theorem; monotonicity is one such hypothesis, and it is absent here.
Check Your Understanding
For each question, track both what happens at a fixed point and the measure of the sets where the indicators equal one.
- For \(r_n=\chi_{[3n,3n+1)}\) on \((\mathbb{R},\mathcal{L},m)\), determine the pointwise limit and compute \(\int r_n\,dm\) for each \(n\).
- For \(s_n=\chi_{[-n,-n+1)}\), prove that \(s_n(x)\to0\) for every fixed \(x\), and calculate its integral.
- For \(E_n=[0,2)\cup[n+2,n+4)\), find the pointwise limit of \(\chi_{E_n}\) and compare its integral with the integrals of \(\chi_{E_n}\).
- At which points do \(f_1=\chi_{[1,2)}\) and \(f_2=\chi_{[2,3)}\) show that this sequence is neither increasing nor decreasing?
- For \(f_n=\chi_{[n,n+1)}\), compute \(m(\{|f_n|>\tfrac12\})\). What does this value imply about convergence in measure to zero?