Let \(f:[0,\infty)\to[0,\infty)\) be continuous and suppose \[ \int_0^\infty f(x)\,dx<\infty. \] Prove that \[ \lim_{n\to\infty}\int_0^n\frac{x f(x)}{n}\,dx=0. \]
Solution
Let \[ I_n\coloneqq\int_0^n\frac{x}{n}f(x)\,dx. \]
1. For every \(\varepsilon>0\), there exists \(A>0\) such that\[ \int_A^\infty f(x)\,dx<\frac{\varepsilon}{2}. \]
The function \(f\) is nonnegative and \[ \int_0^\infty f(x)\,dx<\infty. \] Therefore the tails of this improper integral tend to zero.
2. For \(n>A\),\[ 0\leq I_n \leq \frac{A}{n}\int_0^A f(x)\,dx + \int_A^\infty f(x)\,dx. \]
Split the integral at \(A\): \[ I_n = \int_0^A\frac{x}{n}f(x)\,dx + \int_A^n\frac{x}{n}f(x)\,dx. \] On \([0,A]\) one has \(0\leq x/n\leq A/n\), while on \([A,n]\) one has \(0\leq x/n\leq1\). Since \(f\geq0\), \[ I_n \leq \frac{A}{n}\int_0^Af(x)\,dx + \int_A^nf(x)\,dx, \] and the last integral is at most the tail integral in the claim.
3. For all sufficiently large \(n\),\[ 0\leq I_n<\varepsilon. \]
1. The finite number\[ C_A\coloneqq A\int_0^A f(x)\,dx \]is independent of \(n\). Choose \(n>A\) sufficiently large that\[ \frac{C_A}{n}<\frac{\varepsilon}{2}. \]Then step 2 and step 1 give\[ 0\leq I_n < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \]
4. Therefore\[ \boxed{ \lim_{n\to\infty} \int_0^n\frac{x f(x)}{n}\,dx =0 }. \]
Step 3 is precisely the \(\varepsilon\)-definition of \(I_n\to0\).
5. Q.E.D.
Step 4 is the required limit.