Counterexamples

(The limit of the integrals need not be the integral of the limit)

On \(\RR\) with Lebesgue measure, each of the following sequences converges pointwise to \(0\) while \(\int f_n = 1\) for every \(n\):

• \(f_n \coloneqq n\chi_{[0,1/n]}\), whose mass concentrates near \(0\);

• \(f_n \coloneqq \frac1n\chi_{[0,n]}\), which is uniformly bounded by \(1\) with supports of unbounded measure, so the bounded convergence theorem fails without the hypothesis that the \(f_n\) are supported in one set of finite measure;

• \(f_n \coloneqq \chi_{[n,n+1]}\), whose mass escapes to infinity.

In each case \(\sup_n f_n\) is not integrable, so no integrable function dominates the sequence, and the dominated convergence theorem does not apply. Each sequence also converges almost everywhere without converging in \(L^1\).

(Convergence in $L^p$ does not imply almost-everywhere convergence)

For \(n = 2^k + j\) with \(0\leq j < 2^k\), let \(f_n \coloneqq \chi_{[j2^{-k},\,(j+1)2^{-k}]}\) on \([0,1]\) (the typewriter sequence). Then \(\norm{f_n}_p = 2^{-k/p}\to 0\) for every \(p<\infty\), but for every \(x\in[0,1]\) the sequence \(f_n(x)\) takes the value \(1\) infinitely often and the value \(0\) infinitely often, so it converges at no point. Every sequence converging in \(L^p\) has a subsequence converging almost everywhere; here \(f_{2^k}=\chi_{[0,2^{-k}]}\to 0\) on \((0,1]\).

(A subset of $\RR$ that is not Lebesgue measurable)

The Vitali set, a set of representatives of \(\RR/\QQ\) in \([0,1]\) chosen using the axiom of choice, is not Lebesgue measurable.

(A Lebesgue measurable set that is not Borel)

The Lebesgue \(\sigma\)-algebra is the completion of the Borel \(\sigma\)-algebra. Let \(c\) be the Cantor function and \(\psi(x) \coloneqq x + c(x)\), a homeomorphism \([0,1]\to[0,2]\) that maps the Cantor set \(C\) onto a set of measure \(1\). Choose a non-measurable \(A\subseteq\psi(C)\) and put \(B\coloneqq\psi^{-1}(A)\subseteq C\). Then \(B\) is Lebesgue measurable, being a subset of the null set \(C\), but \(B\) is not Borel, since \(\psi(B)=A\) is the preimage of \(B\) under the continuous map \(\psi^{-1}\) and preimages of Borel sets under continuous maps are Borel.

(A composition of Lebesgue measurable functions need not be measurable)

With \(\psi\), \(A\), and \(B\) as in the construction of a Lebesgue measurable set that is not Borel, \(\chi_B\) is Lebesgue measurable and \(\psi^{-1}\colon[0,2]\to[0,1]\) is continuous, but \(\chi_B\circ\psi^{-1} = \chi_A\) is not Lebesgue measurable. If \(g\) is continuous and \(f\) is Lebesgue measurable, then \(g\circ f\) is Lebesgue measurable.

(A set of positive measure need not contain an interval)

A fat Cantor set, obtained from \([0,1]\) by removing at stage \(n\geq1\) an open middle interval of length \(4^{-n}\) from each of the \(2^{n-1}\) remaining closed intervals, is closed, has measure \(1/2\), and contains no interval.

(Neither of $L^1(\RR)$ and $L^2(\RR)$ contains the other)

The function \(x^{-1/2}\chi_{(0,1)}\) is in \(L^1(\RR)\) and not in \(L^2(\RR)\), and \(x^{-1}\chi_{(1,\infty)}\) is in \(L^2(\RR)\) and not in \(L^1(\RR)\). On a space of finite measure, \(L^q\subseteq L^p\) for \(p<q\).

(The dual of $L^\infty$ is not $L^1$)

The map \(L^1([0,1])\to (L^\infty([0,1]))^*\), \(g\mapsto\qty{f\mapsto\int fg}\), is not surjective: a Hahn–Banach extension \(\varphi\) of \(f\mapsto f(0)\) from \(C([0,1])\subseteq L^\infty([0,1])\) is not of this form, since continuous \(f_n\) with \(0\leq f_n\leq 1\), \(f_n(0)=1\), and \(f_n\to0\) on \((0,1]\) give \(\varphi(f_n)=1\) while \(\int f_ng\to0\) by dominated convergence. For a \(\sigma\)-finite measure, \((L^1)^* \cong L^\infty\).

($L^p$ is a Hilbert space only for $p=2$)

If the measure space contains disjoint sets \(E, F\) of measure \(1\), then \(f\coloneqq\chi_E\) and \(g\coloneqq\chi_F\) satisfy \(\norm{f+g}_p^2+\norm{f-g}_p^2 = 2\cdot 2^{2/p}\) and \(2\norm f_p^2+2\norm g_p^2 = 4\), so the parallelogram law fails unless \(p=2\).

(A pointwise limit of continuous functions need not be continuous)

\(f_n(x)\coloneqq x^n\) on \([0,1]\) converges pointwise to \(\chi_{\theset{1}}\).

(A uniform limit of differentiable functions need not be differentiable)

\(f_n(x)\coloneqq\sqrt{x^2+1/n}\) converges uniformly on \(\RR\) to \(\abs x\), which is not differentiable at \(0\). If \(f_n\) are differentiable on \([a,b]\), \(f_n'\) converges uniformly, and \(f_n(x_0)\) converges for some \(x_0\in[a,b]\), then \(f_n\) converges uniformly to a differentiable \(f\) with \(f' = \lim f_n'\).

(A differentiable function need not be continuously differentiable)

\(f(x)\coloneqq x^2\sin(1/x)\) for \(x\neq0\) and \(f(0)\coloneqq0\) is differentiable on \(\RR\) with \(f'(0)=0\), and \(f'(x) = 2x\sin(1/x)-\cos(1/x)\) has no limit as \(x\to0\).

(A continuous function on a bounded set need not be uniformly continuous)

\(f(x)\coloneqq 1/x\) is continuous on \((0,1)\) and not uniformly continuous: \(\abs{f(1/n)-f(1/(n+1))}=1\) while \(\abs{1/n-1/(n+1)}\to0\). A continuous function on a compact metric space is uniformly continuous.