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An exercise in Real and Complex analysis (exercise 18, page 195,chapter 9)

Show that if a function $f$ on the real line $\mathbb{R}$ satisfies

\begin{equation*} f(x+y)=f(x)+f(y) \end{equation*}

and if $f$ possesses Lebesgue measure, then $f$ is continuous.

I can't do this question. Any advice or hints would be appreciated. Thank you in advance.

hrkrshnn
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Le Hoang Minh
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  • Welcome to Math.SE! Could you please include what you have tried? That way, it is easier for other people to help you. – Hrodelbert May 13 '15 at 13:48
  • Notice that $f(2x)=2f(x) , f(3x)=3f(x) ,f(nx)=nf(x)$ – Someone May 13 '15 at 14:31
  • I know that $f(I)$ is dense in $\Bbb R$ for each open interval $I$ or $f$ is continuous. It's not too hard, but this new statement (new for me) seems much stronger. – ajotatxe May 13 '15 at 14:51
  • You need the following theorem: Lusin's Theorem: Let $f:[0,1]\rightarrow [0,\infty)$ be a nonnegative, measurable function. Suppose $ϵ>0$. Then $∃$ a compact set $K\subset [0,1]$ such that $m(K)>1−ϵ$ ($K$ fills out most of $[0,1]$), and a continuous function $g$ on $[0,1]$ such that $g(x)=f(x)$ if $x∈K$. –  Sep 10 '15 at 10:16
  • @Hrodelbert: I know that too, but I couldn't apply it to prove my exercise. – Le Hoang Minh Sep 16 '15 at 03:35

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