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Let $(X,A,\mu)$ be a measure space. Take a sequence $(f_n)_{n \in \mathbb N}$ of real-valued measurable bounded functions. Suppose $f_n \to f$ uniformly on $X$ and suppose that $\mu(X)<+\infty$. Then $$ \int_X f_n d\mu \to \int_Xf d\mu $$ when $n \to +\infty$.

This exercise is taken from Rudin, Real and Complex Analysis, chapter 1. I do not understand where I should use the hypotesis of boundness of the functions. Indeed, $$ \left \vert \int_X f-f_n d\mu \right\vert \le \int_X \vert f_n - f \vert d\mu \le \int_X \varepsilon d\mu = \varepsilon \mu(X) $$ for $n$ sufficiently big.

Where do I use the boundness of the functions? Do I need it in order to say $\vert \int_X f_n-f d\mu\vert \le \int \vert f-f_n \vert d\mu$? Thanks in advance.

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up vote 4 down vote accepted

Boundedness of the maps and finiteness of the measure space are used to be sure that $\int_Xf_nd\mu$ and $\int_Xfd\mu$ are real numbers.

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Thanks, I understand now. So my proof is coorect, provided I say: "All these quantities are finite because $f_n$ (hence $f$ which is their uniform limit) are bounded". Thanks a lot for your kind and fast help. –  Romeo Nov 10 '12 at 16:23
    
Yes. Actually, it's a sufficient condition to assume boundedness. For example, integrability of all these functions is enough. –  Davide Giraudo Nov 10 '12 at 16:26
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If the functions are not bounded, you may not be able to consider the absolute value of the difference. you might end up with infinity-infinity which is undefined.

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