Let \((f_k)\) be a sequence in \(L^2([0,1])\) satisfying \[ \|f_k\|_2\le M \qquad\text{for every }k. \] Assume \(f_k\to f\) almost everywhere. Prove that \(f\in L^2([0,1])\), that \(\|f\|_2\le M\), and that \[ \int_0^1 f_k(x)\,dx\longrightarrow\int_0^1 f(x)\,dx. \]
Solution
1. The limit belongs to \(L^2\) with the same bound.
Since \(|f_k|^2\to |f|^2\) almost everywhere, Fatou’s lemma gives \[ \int_0^1|f|^2 \le \liminf_{k\to\infty}\int_0^1|f_k|^2 \le M^2. \] Hence \[ f\in L^2([0,1]) \qquad\text{and}\qquad \|f\|_2\le M. \]
2. Prove \(f_k\to f\) in \(L^1\).
Fix \(\varepsilon>0\). Choose \(\delta>0\) such that \[ 2M\sqrt\delta<\frac\varepsilon2. \] By Egorov’s theorem, there is a measurable set \(F\subset[0,1]\) such that \[ m([0,1]\setminus F)<\delta \] and \(f_k\to f\) uniformly on \(F\).
Thus for all sufficiently large \(k\), \[ \int_F|f_k-f| \le m(F)\sup_F|f_k-f| <\frac\varepsilon2. \] On the exceptional set \(E=[0,1]\setminus F\), Cauchy–Schwarz gives \[ \begin{aligned} \int_E|f_k-f| &\le m(E)^{1/2}\|f_k-f\|_2\\ &\le \sqrt\delta\,(\|f_k\|_2+\|f\|_2)\\ &\le 2M\sqrt\delta <\frac\varepsilon2. \end{aligned} \] Therefore \[ \|f_k-f\|_1<\varepsilon \] for all sufficiently large \(k\). Hence \[ f_k\to f\quad\text{in }L^1([0,1]). \]
3. Conclude convergence of the integrals.
Since \[ \left|\int_0^1 f_k-\int_0^1 f\right| \le \int_0^1|f_k-f| =\|f_k-f\|_1, \] Step 2 yields \[ \boxed{ \int_0^1 f_k(x)\,dx\to\int_0^1 f(x)\,dx.} \]