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I want to prove that if $f:\mathbb{R}\to\mathbb{R}$ is Lebesgue-integrable, then for every $\epsilon >0$ there exists $g:\mathbb{R}\to\mathbb{R}$ continuous such that $\displaystyle\int_{\mathbb{R}}|f-g|<\epsilon$ (I will write the integral simply $\displaystyle\int|f-g|$).

I already know how to approximate $f$ by simple functions.

Now, in the following link:

Finding simple, step, and continuous functions to satisfy Lebesgue integral conditions

The author of the chosen answer proves that for the characteristic function $[E ]$ and $\epsilon >0$, there is a continuous function $\phi:\mathbb{R}\to [0,1]$ such that $\displaystyle\int |[ E]-\phi|<\epsilon$, but he uses the regularity which is only valid if $E$ is of finite measure.

Is this still valid if $E$ isn't of finite measure? Or at least, I'm trying to do the case when $E$ is $\sigma$-finite, because I'm interested when $E=\mathbb{R}$.

Any suggestion? Thanks.

Community
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Surtan
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1 Answers1

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Consider $\mathbb{R}=\bigcup_i [-i,+i]$.

1st ingredient:

For all $i\geq 1$, find continuous $g_i$ supported on $[-i,+i]$ s.t. $\int_{[-i,+i]} \left|f-g_i \right| < \frac{1}{i}$.

This is possible thanks to the proof in the link you provided.

2nd ingredient:

Now since $f:\mathbb{R}\rightarrow\mathbb{R}$ is Lebesgue-integrable, $\int_{\mathbb{R}-[-i,+i]}|f|\rightarrow0$ as $i\rightarrow\infty$.

Can you conclude the argument using those ingredients?

Seong Joon Oh
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  • Sorry, but I don't get it. I understand that if $\epsilon i >0$ then exists $i$ such that $\int{\mathbb{R}\setminus{[-i,i]}}|f|<\epsilon i$. And then maybe that would imply $\int{\mathbb{R}}|f-g_i|<\epsilon$, but I don't see what $\epsilon_i$ would be chosen. – Surtan Mar 22 '15 at 18:00
  • Sorry Seong, but I don't know how to tag you. I hope you'll see this. – Surtan Mar 22 '15 at 18:06
  • Let $\epsilon>0$. Let $i$ be big enough such that $i<\frac{\epsilon}{2}$. Use 1st ingredient: pick $g_i$ supported on $[-i,+i]$ which $L^1$-approximates $f$ by $\frac{1}{i}$ (and thus by $\frac{\epsilon}{2}$). – Seong Joon Oh Mar 22 '15 at 20:01
  • Now use ingredient 2 to find $i^\prime$ such that $\int_{\mathbb{R}-[-i^\prime,+i^\prime]}|f| < \frac{\epsilon}{2}$. – Seong Joon Oh Mar 22 '15 at 20:07
  • Then, we define $j=max{i,i^\prime}$ and decompose the integral $\int |f-g_j| = \int_{[-j,+j]}|f-g_j| + \int_{\mathbb{R}-[-j,+j]}|f|$... – Seong Joon Oh Mar 22 '15 at 20:10
  • Hope there was no typo... The LaTex parser is not working atm. – Seong Joon Oh Mar 22 '15 at 20:11
  • I don't get why $\int |f-g_j|=\int_{[-j,+j]|f-g_j|}+\int_{\mathbb{R}\setminus{[-j,+j]}}|f|$...

    Shouldn't it be $\int |f-g_j|=\int_{[-j,+j]|f-g_j|}+\int_{\mathbb{R}\setminus{[-j,+j]}}|f-g_j|$?

     – Surtan Mar 22 '15 at 22:06
  • That's the key point here. That's because $g_j$ is supported on $[-j,+j]$. i.e. $g_j=0$ on $\mathbb{R}-[-j,+j]$. – Seong Joon Oh Mar 22 '15 at 22:07
  • In the link I provided, the autor says that $\phi\restriction_{U^c}=0$, where $U$ is an open set and $E\subseteq U$. In this problem we would take $E=[-j,j]$. But I don't see why $g_j=0$ on $\mathbb{R}\setminus E$ (namely $E^c$). – Surtan Mar 22 '15 at 22:21