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A function that is almost everywhere continuous is in $L_2$; however, the converse might not be true. I couldn't find any example to show this, could you help me with this?

$$L_2= \left \{ f:\mathbb{R}\to\mathbb{R} | \int f^2(x)dx < \infty \right \} $$

Saj_Eda
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    What do you mean by $L_2$? – Michael Albanese Feb 22 '16 at 04:38
  • Think about the nowhere continuous ones. And continuous almost everywhere does not ensure $L^2$-integrability. – Ningxin Feb 22 '16 at 04:39
  • @Wen Are those $L_2$ integrable? – Saj_Eda Feb 22 '16 at 04:41
  • "A function that is almost everywhere continuous is in L2" L2 of what? And why do think this is true? – zhw. Feb 22 '16 at 04:50
  • The space $L_2$ is given in the question – Saj_Eda Feb 22 '16 at 23:26
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    Sara, I think that your critics are worried about the fact that for example the constant function $f(x)=1$ is not in $L^2(\Bbb{R})$ even though it is continuous. This is because the integral $\int_{-\infty}^\infty 1^2,dx=\infty$. It is in $L^2([a,b])$ for any compact interval $[a,b]$ though. You need to be a bit more specific here. – Jyrki Lahtonen Feb 26 '16 at 07:05
  • Another example is $f(x)=1/\sqrt{x}$ with $f(0)=0$. This function is almost everywhere continuous on, say, $[0,1]$, since it is discontinuous only at $0$. However, it is not in $L^2([0,1])$: the integral $\int_0^1 1/x , dx$ is infinite. – Sambo Jan 15 '23 at 20:09

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As is known almost everywhere continuous function is measurable. Hence it's in the $L_2$. Conversely, if we suppose that $f(x) \in L_2$, we may consider that $f(x)^2$ is measurable. So it's sufficient to us to find $f(x)$ such that it's square would be measurable but function itself would not.

let $C$ - some immeasurable set such that $C \subset \mathbb{R}$.

we define $f(x):$ $$f(x) = \left\{\begin{matrix}1, \: x \in C\\-1, \: x \notin C\end{matrix}\right.$$ Obviously, $f(x)$ - immeasurable function, but $f^2(x) = 1$ - easy to see, it's measureable.

artem zholus
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