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In Royden's book, he gives a lemma called the simple approximation lemma which states:

Let $f$ be a measurable real-valued function on $E$. Assume $f$ is bounded on $E$; that is, there is an $M \geq 0$ for which $|F| \leq M$ on $E$. Then for each $\varepsilon > 0$, there are simple functions $\varphi_\varepsilon$ and $\psi_\varepsilon$ defined on $E$ which have the following approximation properties:

$\varphi_\varepsilon \leq f \leq \psi_\varepsilon$ and $0 \leq \psi_\varepsilon - \varphi_\varepsilon < \varepsilon$.

Now, to prove that if $f$ is a bounded function on a measurable set $E$ with finite measure implies integrability, he uses $\frac{1}{n}$ instead of $\varepsilon$; that is, he uses the following inequalities in this proof sketch

Using the simple approximation $\varphi_n \leq f \leq \psi_n$ on $E$ and $0 \leq \psi_n - \varphi_n \leq \frac{1}{n}$ on $E$.

By monotonicity and linearity $0 \leq \int_E \psi_n - \int_E \varphi_n = \int_E [\psi_n - \varphi_n] \leq \frac{1}{n} \cdot m(E)$

etc...

What I'm most curious about if $\varepsilon$ is interchangeable with $\frac{1}{n}$ and why you would use one or the other in proofs. In this case, it is not obvious for me.

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  • There is no real difference here. For any $\epsilon>0$ there is some $n$ such that $\frac{1}{n} < \epsilon$, and vice versa. – copper.hat Nov 05 '13 at 19:19
  • He wants to be able to use sequences, and exploits what copper said to do that. I don't think he needs a sequence here, at least not in what you've reproduced, but that's the conceit of it. – zibadawa timmy Nov 05 '13 at 19:24
  • Another benefit to using $\frac 1n$ in a proof, except for being able to use sequences, is that there will only be countably many terms. If you want to use countable additivity of a measure that can be helpful. See for example: http://math.stackexchange.com/questions/20661/the-sum-of-an-uncountable-number-of-positive-numbers – Samuel Nov 05 '13 at 19:29
  • Thank you all, that clarifies a lot. – user1876508 Nov 05 '13 at 21:35

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