# Prove this inequality of integrals:

Let $(X, M, \mu\ )$ be a measure space. Let $f$ be a positive measurable function with $\int_X f d \mu\ < \infty\ .$ Then for every $\epsilon\ > 0$, there is a set $E \in M$ such that $\mu(E)<\infty$ and $$\int_X f d \mu\ \leq \int_E f d \mu\ + \epsilon$$

This problem appears on Bartle's Elements of Integration and Lebesgue Measure. I couldn't prove it for the life of me. And obviously, $E$ has to be a proper subset of $X$, otherwise, this is just trivial. Any idea?

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You must want some hypothesis on $E$ besides being proper. In an atomless space, you could just remove a point, and if the cardinality of $X$ is finite, this would be false. (E.g., if $X$ is a point, $\mu(X)=1$, and $f=1$.) Perhaps you want $\mu(E)<\infty$? – Jonas Meyer May 19 '11 at 17:56
Hmm.. I'm not sure why you're saying this would be false if $\mu\(X)<\infty$ since by monotonicity of the measure, we automatically have $\mu\(E)<\infty$ when $\mu\(X)<\infty$. – Tasadduk Chowdhury May 19 '11 at 21:44
I am not saying that. I am saying that it will typically be false that you can find a proper subset $E$ of $X$ satisfying the above condition if $X$ has a finite number of points (and no nonempty null sets). Will you please double check your source of this problem? I believe you have either misread it or you have a version with a typo. Please update when you determine which. I'd be happy to help with what the problem most likely is, namely finding such $E$ with $\mu(E)<\infty$. (Of course then $E$ need not be proper, and it is trivial when $\mu(X)<\infty$.) – Jonas Meyer May 19 '11 at 22:04
@Jonas: Hey thanks for reading into this... I did leave out the part saying $\mu(E)<\infty$. – Tasadduk Chowdhury May 19 '11 at 23:20
You're welcome. The part where you say $E$ should be proper is still incorrect. In general, it will not be; the example in my first comment gives such a case. (And anytime $\mu(X)<\infty$, you can just take $E=X$, but this special case need not be explicitly addressed.) – Jonas Meyer May 20 '11 at 0:24

Since the result is trivial if $\int_X {fd\mu } = 0$ or $\mu(X) < \infty$, assume that $\int_X {fd\mu } > 0$ and $\mu(X) = \infty$.

Given $\varepsilon > 0$, it follows from the definition $$\int_X {fd\mu } = \sup \bigg\{ \int_X {sd\mu } :0 \le s \le f,s \; {\rm simple}\bigg \}$$ that there exists a simple function $0 \leq s \leq f$ such that $$\int_X {fd\mu } - \varepsilon \le \int_X {sd\mu }.$$ We can suppose that, for some $n \in \mathbb{N}$, $s$ takes positive values $\alpha_i$, $i=1,\ldots,n$, on some measurable subsets $E_i$ of $X$ with (necessarily) $\mu(E_i) < \infty$ (since $\int_X {sd\mu } < \infty$), and $s$ is zero on the complement of $E_1 \cup \cdots \cup E_n$. Noting that $s \leq f$, we then have $$\int_X {sd\mu } = \int_{E_1 \cup \cdots \cup E_n } {sd\mu } \le \int_{E_1 \cup \cdots \cup E_n } {fd\mu } ,$$ from which the desired result follows with $E = E_1 \cup \cdots \cup E_n$.

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Each set $\displaystyle{A_n=\left\{x:|f(x)|>\frac{1}{n}\right\}}$ has finite measure. Each function $f\cdot\chi_{A_n}$ is dominated by $f$, and $f\cdot\chi_{A_n}\to f$ pointwise ($\chi_{A_n}$ denotes the characteristic function of $A_n$). By Lebesgue's dominated convergence theorem, $\int_{A_n}fd\mu\to \int_X fd\mu$ as $n\to\infty$. Therefore $E$ can be taken to be $A_n$ for sufficiently large $n$.

Here you could also just use monotone convergence, because $(f\cdot \chi_{A_n})$ increases to $f$.

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Jonas, doesn't the result follow from the definition of $\int_X {fd\mu }$ (as a sup of...). – Shai Covo May 19 '11 at 23:59
@Shai: Yeah, good point, it could follow more straightforwardly from the definition than I indicated. However, you must have a different definition in mind than I do if this is immediate from the definition. Nonetheless I'm ready to admit this is overkill. Perhaps you could expand on that in an answer? (Of course, everything follows from the definitions :)) – Jonas Meyer May 20 '11 at 0:06
The idea is to consider $\int_X {fd\mu } - \varepsilon \le \int_X {sd\mu } \le \int_{E_1 \cup \cdots \cup E_n } {fd\mu }$ where $\mu (E_1 \cup \cdots \cup E_n ) < \infty$. – Shai Covo May 20 '11 at 0:12
Above $s$ is a simple (nonnegative) function taking positive values on $E_1,\ldots,E_n$ with (necessarily) $\mu(E_i) < \infty$. – Shai Covo May 20 '11 at 0:16
@Shai: Yeah, that's nice. Why not answer? I also like the monotone convergence way of thinking of it, and I like dominated convergence because it would also quickly give $E$ with $\mu(E)<\infty$ and $|\int_Xfd\mu - \int_Efd\mu|<\varepsilon$ regardless of sign. This could all be reduced to the definition, and admittedly, the not-necessarily-nonnegative cases would involve many of the same considerations as working from the definition, but I make no effort to be minimalist. Thank you for your comments, and for pointing out a more optimal explanation. – Jonas Meyer May 20 '11 at 0:17