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I have to prove that a function is either Lebesgue integrable or to give counterexample. In particular, the problem is related to almost direct application of Dominated Convergence theorem.

Prove or disprove $$\lim _{n\rightarrow \infty} \int_{[0,1]}\exp\left(\frac{x^2}{n}\right)dx = \int_{[0,1]} \lim _{n\rightarrow \infty} \exp\left(\frac{x^2}{n}\right)dx.$$

I managed to formulate the following proof:

Looking at the function $f(x) = \exp(\frac{x^2}{n}), x \in [0,1], n \geq 1$. It is obvious that it is dominated by the function $g(x) = \exp(x^2)$, since $\exp(\frac{x^2}{n}) \leq \exp(x^2)$. Moreover, $\int _{[0,1]} e^{x^2} dx< +\infty$. From the above-stated theorem it follows that $f(x)$ is Lebesgue Integrable and $\lim _{n\rightarrow \infty} \int_{[0,1]}\exp(\frac{x^2}{n})dx = \int_{[0,1]} \lim _{n\rightarrow \infty} \exp(\frac{x^2}{n})dx$.

My question here are two. First, I am not sure about the dominance of $e^{x^2}$ on that interval. Second, since the function is dominated, it follows that it is Lebesgue integrable ($g(x)$ is integrable). Does it follow that we can interchange the limits as done in the statement?

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    $\begingroup$ Since $\exp$ is (strictly) increasing, the domination is correct. Further, $x\mapsto \exp(x^2)$ is continuous on the compact interval $[0,1]$, and hence it is bounded by a constant $C$, so $\int_0^1 \exp(x^2)dx\le C m([0,1])=C<\infty$ $\endgroup$ – Reveillark May 7 '17 at 18:39
  • $\begingroup$ @S.19LaBG Isn't that what the DCT says? $\endgroup$ – zhw. May 7 '17 at 18:58

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