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I'm trying to find an example of a measure $\mu$ and a nonnegative finite-valued Borel measurable function g such that the measure $\lambda$ defined by $$\lambda(A) = \int_Ag d\mu$$ is not $\sigma$-finite. I've played around with various counting measures on various spaces, but can't seem to quite get it. I've read in "Counterexamples in Analysis" that there exists a function which is nonnegative, finite and measurable, but whose integral is $\infty$ on every open interval (8.29, page 105). Can $\lambda$ be $\sigma$-finite even for such a function?

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If the measure $\mu$ is $\sigma$-finite, then the measure $\lambda$ will be $\sigma$-finite. Indeed, let $\left(B_j\right)_{ j\geqslant 1}$ be a collection of pairwise disjoint sets which have a finite $\mu$-measure. Define $$A_{j,l} := B_j\cap\left\{2^l\leqslant g\lt 2^{l+1} \right\} , \quad j\geqslant 1,l\in\mathbb Z $$ and $C_0 :=\{g=0\}$. Then the collection $\left\{C_0, A_{j,l}, j\geqslant 1,l\in\mathbb Z \right\}$ forms a countable partition of the measure space and each set has a finite $\lambda$-measure (that of $C_0$ is $0$, that of $A_{j,l}$ does not exceed $2^{l+1}\mu\left(B_j\right)$).

On the other hand, when the measure $\mu$ is not $\sigma$-finite, just choose $g=1$.

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