Approximation to identity, but average value is zero This is a problem from Real Analysis(Stein, Shakarchi).

Suppose $\{K_\delta\}$ is a family of kernels that satisfy the following:  (i)
$|K_\delta(x)|\leq A\delta^{-d}$ for all $\delta>0$.  (ii) $|K_\delta(x)|\leq \frac{A\delta}{|x|^{d+1}}$ for all $\delta>0$  (iii) $\int_{-\infty}^{\infty}K_\delta(x)dx=0$ for all $\delta>0$  Thus, $K_\delta$ satisfies conditions (i) and (ii) of approximations to identity, but the average value of $K_\delta$ is $0$ instead of $1$. Show that if $f$ is integrable on $\mathbb{R}^d$, then
$$(f * K_\delta)(x) \to 0 \quad\text{for a.e x , as}\quad \delta \to 0$$

First of all, I tried to mimic the proof for the approximation to the identity with the integral value being zero. The book used the fact that the function
$$A(r)=\frac{1}{r^d}\int_{|y|\leq r}|f(x-y)-f(x)|dy$$
is continuous function of $r$, and that $A(r) \to 0$, as $r \to 0$. As (i) and (ii) are both same in approximation to the identity, and in the problem, it leads me to similar steps. But, the problem is that I don't know how to use the condition (iii). If $\int_{\mathbb{R}}|K_\delta(x)|=0$, then I can conclude that $K_\delta(x)=0$ a.e $x$, but this is not the case. Also, using the triangular identity for integrals does not give me any information. How do I need to interpret condition (iii)?
 A: You are right, the entire argument is essentially the same, the condition $3$ only changes what you have actually proved.
In both cases you use the lemma on $A(r)$ and properties $(i)$ and $(ii)$ to show
$$\lim_{\delta\to 0}\left\vert\int (f(x-y)-f(x))K_{\delta}(y)dy\right\vert\to 0\ , $$
for a.e. $x$. The difference is what the term in absolute value means. If $\int K_{\delta}(y)dy=1$ then
$$\begin{align}\int (f(x-y)-f(x))K_{\delta}(y)dy&= \int f(x-y)K_{\delta}(y)dy-\int f(x)K_{\delta}(y)dy\\
&=(f\ast K_{\delta})(x)-f(x)\int K_{\delta}(y)dy\\&= f\ast K_{\delta}(x)-f(x)\end{align}$$
and so the limit above going to $0$ means
$$\lim_{\delta\to 0}(f\ast K_{\delta})(x)\to f(x)\ .$$
However, if $\int K_{\delta}(y)dy=0$ then the integral in the absolute value is
$$\begin{align}\int (f(x-y)-f(x))K_{\delta}(y)dy &= \int f(x-y)K_{\delta}(y)dx -\int f(x)K_{\delta}(y)dy\\
&=(f\ast K_{\delta})(x)-f(x)\int K_{\delta}(y)dy\\
&=(f\ast K_{\delta})(x)\ .\end{align}$$
And so you have proved
$$\lim_{\delta\to 0} (f\ast K_{\delta})(x)\to 0\ , $$
for a.e. $x$.
