Claim in Stein and Shakarchi Real Analysis, made stronger In Stein and Shakarchi, Real Analysis textbook they claim that 
"The main use of good kernels was that whenever $f$ is bounded then $(f*K_\delta)(x)\rightarrow f(x)$ as $\delta\rightarrow 0$ at every point of continuity of $f$". They define good kernels as a family of Lebesgue integrable functions on $\mathbb{R}^d$ that satisfy $\int K_\delta(x)dx=1$, $\int |K_\delta(x)|dx\leq A$ and for all $\eta>0$ $\int_{|x|\geq\eta} |K_\delta(x)|dx\rightarrow 0$ as $\delta\rightarrow 0$. Now it seems to me like as long as $f$ is bounded and integrable we will always have $(f*K_\delta)(x)\rightarrow f(x)$ as $\delta\rightarrow 0$ even if $f$ is not continuous at $x$. Here's why, according to me. Suppose $|f(x)|\leq M$ for all $x\in \mathbb{R}^d$ 
Given $\epsilon>0$ we have 
$|(f*K_\delta)(x)-f(x)|=|\int f(x-y)K_\delta(y) dy -f(x)|=|\int((f(x-y)-f(x))K_\delta(y)dy|=|\int_{|y|\geq\eta}((f(x-y)-f(x))K_\delta(y)dy + \int_{|y|<\eta}((f(x-y)-f(x))K_\delta(y)dy|\leq |\int_{|y|\geq\eta}((f(x-y)-f(x))K_\delta(y)dy| + |\int_{|y|<\eta}((f(x-y)-f(x))K_\delta(y)dy| \leq \int_{|y|\geq\eta}|((f(x-y)-f(x))K_\delta(y)|dy + \int_{|y|<\eta}|((f(x-y)-f(x))K_\delta(y)|dy=\int_{|y|\geq\eta}|((f(x-y)-f(x))|\cdot|K_\delta(y)|dy + \int_{|y|<\eta}|((f(x-y)-f(x))|\cdot|K_\delta(y)|dy\leq 2M\int_{|y|\geq\eta}|K_\delta(y)|dy + 2M\int_{|y|<\eta}|K_\delta(y)|dy$. 
Now the first sum, by the third property of good kernels, can be made smaller than $\frac{\epsilon}{4M}$. Notice $\eta$ so far has been arbitrary. Since $K_\delta$ is integrable we can choose $\eta$ very small s.t. the second sum is smaller than $\frac{\epsilon}{4M}$. Hence for $\delta$ smaller than the one we chose in the first case, the sum is smaller than $\epsilon$. 
Now I'm wondering if my argument is valid. Seems to me that Stein  and Shakarchi claim it to hold for all $x$ at which $f$ is continuous but, if $f$ is bounded then we could get rid of this assumption. Am I wrong in this conclusion?

Think about the function $f(x) = 0$ except $f(0)=1$.  Convolving against a good kernel will give you the spot on $0$ function for every $\delta$.  Do you see where your proof breaks down?
