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True or False: Is there a measurable function $f: \mathbb{R} \to (0,\infty)$ such that $\int \chi_I f d\lambda = \infty$ for every nonempty open interval $I \subset \mathbb{R}$ where $\lambda$ represents Lebesgue Measure on $\mathbb{R}$?

I am pretty sure there is such a function as I am not able to prove the statement. However, I cannot think of an example.

Mike
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    Here's a guess / half idea: Every open interval contains some diadic interval of the form $[i/2^k, (i+1)/2^k]$ for some $k$ and $i$. For each $i$ and $k$, you can find a measurable function like that on each such interval using a translated version of $1/x$ (except set to be $0$ at the endpoint where it would be infinity). Then just sum all of these up , weighted by $\frac{1}{2^k}$. Perhaps something like that would work? These integrate to infinity on each open. You'd have to futz around with the weights to make sure that these sums are all finite, and I'm guessing that is possible to do. – Elle Najt May 19 '20 at 17:30
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    In a similar vein, I thought about $\sum_{n\ge0}\frac{1}{2^n}\frac{1}{|x-r_n|}$, where $r_n$ is an enumeration of the rationals, but one needs more work to ensure finiteness of the sum. – Thorgott May 19 '20 at 17:36
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    We may consider $$\sum_{k=0}^{\infty}\sum_{n\in\mathbb{Z}}\frac{1}{2^n\left|2^k x-n\right|}.$$ This is along the same line as the above comments. – Sangchul Lee May 19 '20 at 17:39
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    https://math.stackexchange.com/questions/24413/is-there-a-function-with-infinite-integral-on-every-interval – Matias Heikkilä May 19 '20 at 19:51

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