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The problem you pose is the weak version of a classical result in harmonic analysis, which does not require the continuity nor the compactness of the support of the datum $f$: in order to prove this stronger result, first note that  \begin{align} \int\limits_{\Bbb R} f\big(l(t)\big)\mathrm{d}t=0\:\text{ for every}&\text{ straight line } l\subsetneq\Bbb ...
You can easily check the following proposition. Let $E\subset \mathbb R^N$ be a measurabe set. If $f$ and $g$ are two functions defined on $E$ satisfying $f=g$ for a.e.$x\in E$, then $f$ is measurable if and only if $g$ is measurable. In this problem, we can thus assume $f\neq0$ on $I$. Since $\frac1x$ is a continuous function on $\mathbb R- \{0\}$, $\... 2 Suppose there exists$p\in\mathbb{R}^2$with$f(p)\neq 0$, then WLOG (by replacing$f$by$-f$if necessary) we might assume$f(p)>0$. But then by continuity of$f$there exist$\varepsilon>0$such that$f(x)>f(p)/2$for all$x$with$\|x-p\|\leq\varepsilon$, which then gives that the integral of$f$over every line contained in$B(p,\varepsilon)$... 2 You may find this useful for real analysis, and this for functional analysis. Moreover, you can find more video courses here about some other topics in mathematics. If Chinese is OK to you, this course from Shanghai Jiao Tong University is acessible. 1 Let$\mu (\emptyset)=0$and$\mu (A)=\infty$for every non-empty set$A$. Then any sequence$(f_n)$of measurable functions converges in measure to any measurable function$f$! 1 Let$D=A(B(0,1))$. Then$E_{\eta}=\{x:d(x,D) <\eta \}$. We know that$T(B(0,r))\subset \{x:rD)<\eta r\}$. [ By linearity of$A$,$rD$is same as$A(B(0,r))$]. Now$x \in E_{\eta}$iff$rx \in \{x:rD)<\eta r\}$. [This follows from the fact that$\|ry\|=r\|y\|$]. Hence$m(T(B(0,r)) \leq m(rE_{\eta})=r^{k}m(E_{\eta})<r^{k}\epsilon\$. Notation used: ...