Questions about theorem 2.1 and lemma 2.2 in chapter 3 of Stein's Real Analysis I have a difficulty dealing with some proofs in theorem 2.1 and lemma 2.2 in chapter 3 of Stein's Real Analysis. The theorem 2.1 and lemma 2.2 are as follows. For reference, pictures containing the theorems and lemma are posted as well.
Theorem 2.1: {$K_\delta$} is an approximation to the identity and $f$ is integrable on $R^d$. Then, $ (f*K_\delta)(x) → f(x)$ as $\delta → 0 $ for every $x$ in the Lebesgue set of $f$. (In particular, the limit holds for almost everywhere $x$)  Theorem 2.1
Lemma 2.2: $f$ is integrable on $R^d$ and $x$ is a point of Lebesgue set of $f$. Let $A(r) = \frac{1}{r^d}\int_{|y|<r}|f(x-y) - f(x)|dy$, whenever $r > 0$. Then, 1) $A(r)$ is a continuous function of $r > 0$, 2) $\lim_{r\to 0}A(r) = 0$, and 3) $A(r)$ is bounded.  Lemma 2.2
The Stein's textbook says the continuity of $A(r)$ follows by invoking the "absolute continuity". Also, the fact that $x$ belongs to the Lebesgue set of $f$ and the measure of a ball of radius $r$ is $v_dr^d$ make the $A(r)$ tend to $0$.  proof of lemma 2.2
This is where I'm stuck to. I can't relate the "absolute continuity" to the countinuity of $A(r)$. Also, I'm not sure how to use the property "$x$ belongs to the Lebesgue set of $f$, and the measure of a ball of radius $r$ is $v_dr^d$ " to show the convergence of $A(r)$.
For reference, let me type the definition of "absolute continuity" and the Lebesgue set of $f$ by Stein.
Absolute Continuity: Suppose $f$ is integrable on $R^d$. Then, for every $\epsilon > 0$ there is a $\delta > 0$ such that $\int_E|f| < \epsilon$ whenever $m(E) < \delta$. absolute continuity
Lebesgue set of $f$: If $f$ is locally integrable on $R^d$, the Lebesgue set of $f$ consists of all points $x ∈ R^d$ for which $f(x)$ is finite and $\lim_{m(B)\to 0}\frac{1}{m(B)}\int_B|f(y)-f(x)|dy = 0$ (where $x ∈ B$). Lebesgue set of $f$
Any help will be appreciated, especially explanation based on Real Analysis would be grateful (bcz I've only studied undergraduate level of Analysis (Rudin's PMA) and just a beginner of Stein's Real Analysis).
 A: The absolute continuity is used because the difference between the evaluations of the function $A$ at two very close radii is an integral over the difference set between two balls. That is,
$$\int_{|y|< r_1}|f(x-y)-f(x)|dy - \int_{|y|< r_2}|f(x-y)-f(x)|dy= \int_{r_1<|y|< r_2}|f(x-y)-f(x)|dy $$
assuming $0<r_1<r_2$. So you are integrating an integrable function over the set $\{y|r_1<|y|< r_2\}$, whose measure goes to zero as $r_1$ approaches $r_2$ or viceversa. So, $A(r_1)-A(r_2)$ has to go to zero by absolute continuity of the function $|f(x-y)-f(x)|$ (the factor $\frac{1}{r^d}$ does not make much of a difference since it is continuous). In any case, one could use dominated convergence theorem here as well.
Second. You simply check that the thing you want to prove is a special case of the definiton of Lebesgue point. You can make a change of variables and rewrite the integral as
$$ \frac{1}{r^d}\int_{|y|< r}|f(x-y)-f(x)|dy= \frac{1}{r^d}\int_{|z-x|< r}|f(z)-f(x)|dz, $$
and note that the set $\{z||z-x|< r\}$ is a ball that contains $x$, and the factor ${r^d}$ is exactly the size of the ball, up to a fixed multiplicative constant. Therefore, by the definition of Lebesgue point, it has to go to zero as $r$ approaches zero.

Edit: clarifying some details.
So first, for the continuity of $A$ is enough to show the continuity of the function $F:r\mapsto {r^d}A(r)$, because $A(r)=F(r)\frac{1}{r^d}$ and $\frac{1}{r^d}$ is a continuous function (and the product of two continuous fct. is continuous). So, we consider $F$ instead of $A$ (i.e., we throw away the factor $\frac{1}{r^d}$).
Now, fix $r>0$. Fix $\epsilon>0$. By absolute continuity of $g(y):=|f(x-y)-f(x)|$, there exists $\delta$ such that the integral of $g$ over a set $E$ is less than $\epsilon$ if $|E|<\delta$. Now note as before that, for a given $s>0$, we have
$$ |F(r)-F(s)|=\int_{E}g(y)dy, $$
where $E=\{ r<|y|< s\}$ if $s>r$ and $E=\{ s<|y|< r\}$ if $s<r$. In both cases, we have $|E|=C|r^d-s^d|$ for some constant $C$ (just subtract the volumes of the two balls), so that it is less than $\delta$ when $|s-r|<\delta’$, where $\delta’>0$ depends only on $\delta$ and $r$ (you can compute it yourself, but it is not important at all, you simply lnow it exists because of the continuity of the function $s\mapsto C|r^d-s^d|$). So… combining everything together, if $|r-s|<\delta’$, then $|F(r)-F(s)|<\epsilon$, which is the definition of continuity since $\epsilon>0$ is arbitrary. So $f$ is continuous at the point $r$.
When doing analysis, this kind of arguments are used all the time, so you first have to learn to do this by hand and then with some experience you see it immediately and you don’t need to write the details every time to convince yourself. Writing all the details, though, is still quite long as you can see XD
