A weaker form of Lebesgue's differentiation theorem in $\Bbb R ^n$

Question

If $f : \Bbb R ^n \to \Bbb C$ is locally-integrable then Lebesgue's differentiation theorem says that $$\lim \limits _{r \to 0} \frac 1 {\lambda \big( B(x, r) \big)} \int \limits _{B(x, r)} f \Bbb d \lambda = f(x) \tag{*}$$ almost everywhere.

What happens if I want to study the set of those points $x$ where $\lim \limits _{r \to 0} \frac 1 {\lambda \big( B(x, r) \big)} \int \limits _{B(x, r)} f \Bbb d \lambda$ simply exists, without having to be equal to $f(x)$? Is it equal to the set of the points where $(*)$ holds, or can it be larger?

Edit: The above question has been answered below, but I would like to complement it with the following closely related one: are there functions $f$ and points $x$ in which the above limit does not exist? I would like to understand whether in general this limit exists everywhere or only almost everywhere.

Answer 1

Yes, the limit can fail to exist. Say $x_n=2^{-n}$, $y_n=(3/4)x_n$. Let $I_n=[y_n,x_n]$ and set $f=\sum\chi_{I_n}$. Then there exist $a$ and $b$ with $a\ne b$ so that $\frac1x\int_0^x f$ equals $a$ for $x=x_n$ and $b$ for $x=y_n$.

Answer 2

Take any cotinuous $f$ defined on $\mathbb{R}^n$, and any set $E\subset\mathbb{R}^n$ of measure zero.

Define $$g(x) = \begin{cases} f(x) & \text{if}\quad x\in\mathbb{R}^n\setminus E,\\ f(x) + 1, & \text{if}\quad x\in E. \end{cases}$$

So, the limit of $(*)$ for $f$ exists and equals to $f(x)$ everywhere on $\mathbb{R}^n$, however for $g$ the limit exists everywhere, but equals to $g(x)$ only out of $E$.