Let $f$ and $g$ be two real-valued measurable functions (w.r.t. Lebesgue measure) defined on $\mathbb{R}$ such that $$\|f\|_{L^{\infty}(D(x,1))}\leq g(x),\forall x\in\mathbb{R}.$$ Here $D(x,1) = \{y\in\mathbb{R}: |x-y|<1\}$.
Is it true that $f(x)\leq g(x)$ for almost every $x\in\mathbb{R}$?
Any suggestion would be really helpful to me.
If you want to avoid the Lebesgue differentiation theorem, assume without loss of generality that $f$ is non-negative. Since $f$ is measurable, you can find an increasing sequence of simple functions $s_n$ such that $0\le s_n\le f$ and $s_n$ converges to $f$ pointwise. Now fix $n$ and assume that your simple function is $$s_n=\sum_{k=1}^lc_k\chi_{E_k}$$ If the inequality $s_n(x)\le g(x)$ a.e. fails, you can find $k$ and a set $E\subset E_k$ with positive measure such that $c_k>g(x)$ for all $x\in E$. Now partition the real line into intervals of length $1/8$. If the intersection of $E$ with each interval has measure zero, then $E$ would have measure zero. So one of this intervals must intersect $E$ on a set of measure non zero. Hence, without loss of generality you can assume that $E$ is contained in some interval of length $1/8$. Take $x\in E$. Then $E$ is contained in $D(x, 1)$. But then $$\| s_n\|_{L^\infty (D(x, 1))}\ge c_k>g(x),$$ which is a contradiction. Hence $s_n(x)\le g(x)$ a.e. and letting $n\to\infty$ you get the same inequality for $f$.