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Let $\{f_n \}_{n \geq 1}$ be a sequence of non-negative continuous functions defined on $[0,1].$ Assume that $f_n (x) \to f(x)$ for all $x \in [0,1].$ Let $f_n (x) \leq f(x),$ for all $x \in [0,1].$ Show that $$\displaystyle {\lim\limits_{n \to \infty} \int_{0}^{1} f_n (x)\ dx = \int_{0}^{1} f(x)\ dx}.$$

I am trying to apply DCT here but can't able to do that as I don't know whether or not $f$ is Lebesgue integrable. Since we are in a finite measure space so if we can prove that $f$ is bounded then we are through. But I don't think it's true. Here's a counter-example. Consider the sequence of non-negative continuous functions on $[0,1]$ defined by $$ f_n(x) = \left\{ \begin{array}{ll} n^2x, & \quad 0 \leq x \leq \dfrac 1 n; \\ \frac 1 x, & \quad \dfrac 1 n \lt x \leq 1. \end{array} \right. $$

This sequence of functions $\{f_n\}_{n \geq 1}$ converges pointwise to the function $f$ on $[0,1]$ defined by $$ f(x) = \left\{ \begin{array}{ll} 0, & \quad x = 0; \\ \dfrac 1 x, & \quad 0 \lt x \leq 1. \end{array} \right. $$

The function $f$ is clearly unbounded near the point $0.$ So I cannot apply DCT here. So how do I argue for such cases? Any help or suggestion in this regard will be highly appreciated.

Thanks in advance.

Anacardium
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  • @0-thUser Sumanta how do I ensure that the sequence of functions ${f_n }{n \geq 1}$ is uniformly bounded? In my example you can see the sequence of functions is actually not uniformly bounded because $\sup\limits{x \in [0,1]} f_n (x) = n \to \infty.$ – Anacardium Oct 10 '20 at 17:15
  • you should take a look at https://math.stackexchange.com/questions/2713020/why-is-the-monotone-convergence-theorem-more-famous-than-its-stronger-cousin – Paulo Oct 10 '20 at 17:18
  • @0-thUser this question appered in the NBHM PhD entrance $2018.$ So do I assume here that $f$ is integrable? – Anacardium Oct 10 '20 at 17:22
  • I think so, for the notation $\int_0^1 f$. Also, consider $f_n(x)=\frac{1}{x+\frac{1}{n}}$ on $[0,1]$ – Sumanta Oct 10 '20 at 17:24
  • Are you allowed to use Fatou Lemma? From there you can easily conclude – Kolmo Oct 10 '20 at 18:54
  • @Kolmo I don't think that Fatou's lemma is anyway helpful here. Look at my counter-example. If you apply Fatou's lemma there you will find $$\int_{0}^{1} f(x)\ dx \leq \liminf\limits_{n \to \infty} \int_{0}^{1} f_n(x)\ dx = \lim\limits_{n \to \infty} \left (\dfrac 1 2 - \ln \left (\dfrac 1 n \right ) \right ) = \infty.$$ Is it of any importance? – Anacardium Oct 10 '20 at 19:02
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    Well, if f is integrable, you said before, you can use DCT and you are done. If not, your first inequality have LHS equal infinite and therefore the sequence of integrals diverges too. You could also say $$ \int f \leq \liminf \int f_n \leq \limsup \int f_n \leq \int f $$ as $f_n \leq f$. – Kolmo Oct 10 '20 at 19:24
  • @Kolmo very nice remark. Somehow I missed that part. Thanks for pointing that out. – Anacardium Oct 10 '20 at 19:28

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