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While there is a proof given in the book by Richard Bass and I am able to understand it, because I am not able to see why the following is wrong means I am still lacking in understanding:

Given: $\{f_n\}_n$ is an increasing sequence and $lim_{n \to \infty } f_n(x) = f(x), \forall x.$

To Prove: $\int {f_n} d\mu \to \int {f} d\mu $

Attempt: $ |\int (f_n - f)|=|\int f_n - \int f| \leq \int |f_n - f|$, since $ lim_{n \to \infty} f_n(x) = f(x),$

$|f_n - f| \leq \epsilon,$ as $ n \to \infty $. And so, $|\int f_n - \int f| d\mu \leq \epsilon \int d\mu.$

Can someone please give me a hint where I am doing it wrong?

Kevin
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    The problem is that you are using uniform convergence of $f_n$ by estimating $|f_n(x) - f(x)|$ with a single $\epsilon$ on the entire space. As far as I understand, uniform convergence is not given in this theorem. – Denis Gorodkov Jul 13 '21 at 01:18

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In the step where you say $|f_n-f|\leq \epsilon$, what you're actually assuming is uniform convergence, not just pointwise convergence (because for pointwise convergence the $\epsilon$ will depend on the point $x$). Also in your last step, $\int d\mu$ could be infinite which would then make your estimate useless. Also, nowhere in your proof do you use the monotonicity so you already know there's an error. Therefore, your proof is not a proof of the MCT, but upon a careful rereading, it is a proof of the following simpler theorem:

If a sequence $\{f_n\}$ of integrable functions on a finite measure space converges uniformly to $f$, then $f$ is also integrable and $f_n$ converges to $f$ in $L^1(\mu)$ (and thus $\int f_n\, d\mu \to \int f\, d\mu$).

peek-a-boo
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