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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?

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  • $\begingroup$ 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 $m(\{0\})=0$. $\endgroup$
    – Mittens
    Jun 13, 2021 at 14:44

1 Answer 1

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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 $0$.

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