# Lebesgue integral calculation.

Let $$f_k : [0,1] \to \mathbb{R}$$ given by $$f_k(x):=\dfrac{kx^2+1}{(x^2+1)^k}$$.

Then, calculate the limit of Lebesgue integral $$\displaystyle\lim_{k\to \infty} \int_0^1 f_k(x) dx$$ by using dominated convergence theorem.

My arguement is as follows.

Since $$kx^2+1 \leqq 1+kx^2+\dfrac{k(k-1)}{2} x^4+ \cdots +x^{2k}= (1+x^2)^k$$, $$|f_k(x)| \leqq 1.$$ That is, $$f_k$$ is bounded.

$$m^* ([0,1])=1 < \infty.$$

And $$\displaystyle\lim_{k \to \infty} f_k(x)=0$$ if $$x \in (0,1]$$, $$\displaystyle\lim_{k \to \infty} f_k(x)=1$$ if $$x \in \{0 \}.$$ Thus $$\displaystyle\lim_{k \to \infty} f_k(x)= \chi_{\{0\}} (x)$$.

From the dominated convergence theorem, $$\displaystyle\lim_{k\to \infty} \int_0^1 f_k(x) dx= \displaystyle \int_0^1 \lim_{k\to \infty} f_k(x) dx=\int_0^1 \chi_{\{0\}} (x) dx.$$

I'm stacked here. How can I calculate $$\displaystyle\int_0^1 \chi_{\{0\}} (x) dx$$? And is my arguement correct?

• The set $\{0\}$ has Lebesgue measure $0$. Notice that for any measure $\mu$, $\int\mathbb{1}_A(x)\,\mu(dx)mu(A)$. In your problem, you have $m(\{0\})=0$. Jun 13, 2021 at 14:44

Yes, your argument is correct and you are pretty much done. The last integral's calculation is straightforward, since: $$\int_\mathbb{R} \chi_A(x)\text{d}x=m(A)\implies \int_0^1 \chi_{\{0\}}(x)\text{d}x=m(\{0\})=0$$ And the answer is $$0$$.