Convergence of Lebesgue integral of an indicator function

Question

Let $f_n(x) = - \frac{1}{n} \mathbb 1_{[0,n]} (x) $ for $n \in \mathbb N$.

(The 1 should be the indicator function.)

• The sequence converges to $-n-1$, right?
• What about the Lebesgue integral of $f_n$? Can anybody show me if it converges, and if yes, to which value?

Answer 1

If $x<0$, then $f_n(x)=0$ for all $n$. If $x \ge 0$, then there is $N \in \mathbb N$ such that $x \in [0,n]$ for all $n \ge N$, hence $f_n(x)= - \frac{1}{n}$ for all $n \ge N$.

Conclusion: $f_n$ converges pointwise to $0$.

Furthermore: $ \int_{\mathbb R}f_n(x) dx=- \frac{1}{n}\int_{\mathbb R}1_{[0,n}(x) dx=- \frac{1}{n} \lambda([0,n])=- \frac{1}{n}n=-1$,

where $\lambda$ is the Lebesgue measure.

Answer 2

$$\int f_n\;d\lambda=\int-\frac1n\mathbf1_{[0,n]}\;d\lambda=-\frac1n\int\mathbf1_{[0,n]}\;d\lambda=-\frac1n\lambda([0,n])=-\frac1n\cdot n=-1$$ So it converges to $-1$.

Further note that $f_n\to0$ pointwise, and (off course) $\int0\;d\lambda=0$.

So in this case $\lim_{n\to\infty}f_n$ and $\int\lim_{n\to\infty}f_n\;d\lambda$ and $\lim_{n\to\infty}\int f_n\;d\lambda$ are properly defined, but this with:$$\int\lim_{n\to\infty}f_n\;d\lambda\neq\lim_{n\to\infty}\int f_n\;d\lambda$$

Answer 3

$$f_n \to0$$

$$\int_{\mathbb{R}}\frac{-1}{n}1_{[0,n]}(x) \lambda (dx) = \frac{-1}{n}\lambda([0,n]) = -1 \neq \int_{\mathbb{R}} 0 d\lambda$$

so we have an example where we cannot interchange limit and integral.