Assume that \(f_1, f_2, \ldots\) is a sequence of positive continuous functions defined on \([0,1]\) with
\[f(x) = \lim_{n \to \infty} f_n(x) \text{ for every } x \in [0,1]\]
and
\[\int_0^1 f_n(x) \, dx = 1.\]
Is it always true that \(\int_0^1 f(x) \, dx \leq 1\)? Provide a proof if it is true or provide a counterexample if it is false.
Is it always true that \(\int_0^1 f(x) \, dx \geq 1\)? Provide a proof if it is true or provide a counterexample if it is false.
Solution
1. (a) Yes: \(\int_0^1f\le1\).
The \(f_n\) are nonnegative and measurable, and \(f=\liminf_nf_n\) pointwise, so Fatou’s lemma gives \(\int_0^1f\le\liminf_n\int_0^1f_n=1\).
2. (b) No: \(g_n\da\bigl(1-\frac1n\bigr)T_n+\frac1n\), with \(T_n\) the tent of height \(2n\) on \([0,\frac1n]\), is a counterexample.
Let \(T_n(x)=4n^2x\) on \([0,\frac1{2n}]\), \(T_n(x)=4n-4n^2x\) on \([\frac1{2n},\frac1n]\), and \(T_n=0\) on \([\frac1n,1]\); it is continuous with \(\int_0^1T_n=\frac12\cdot\frac1n\cdot2n=1\). So \(g_n\) is continuous, \(g_n\ge\frac1n>0\), and \(\int_0^1g_n=\bigl(1-\frac1n\bigr)+\frac1n=1\). For \(x\in(0,1]\), \(T_n(x)=0\) once \(n>1/x\), and \(T_n(0)=0\), so \(g_n(x)=\frac1n\to0\) for every \(x\). The limit is \(f=0\), with \(\int_0^1f=0<1\).
3. Q.E.D.
Steps 1 and 2 answer parts (a) and (b).