The task is to show that
$$\lim\limits_{n \to \infty} \frac{n}{2}\int_{\mathbb{R}} e^{-n|t|} f(t)dt = 2 $$ where f is bounded and piecewise continuous on $\mathbb{R}$, with $\lim\limits_{t \to 0^+} f(t) = 1$ and $\lim\limits_{t\to 0^-} f(t) = 3$. K$_n$(t) = $\frac{n}{2}e^{-n|t|}$ is a positive summation kernel.
We know that the effect of integrating a function with a positive summation kernel over a symmetric interval is to pick out the mean value of the function at the point t=0, so the limit makes sense. I just don't know how to prove it.
My approach is to show that $\lim\limits_{n\to \infty} \left( \int_{\mathbb{R}} e^{-n|t|} f(t)dt - 2 \right) = 0 $. I first show that $\lim\limits_{n\to \infty}|\left( \int_{\mathbb{R}} e^{-n|t|} f(t)dt - 2 \right) = 0 | = 0$, cause then I can use the triangle inequality and the fact that f is bounded. (What made use the absolute value is honestly just the fact that it was used in the book I'm using in the proof of that result in the case where f is continuous at the origin. In that proof they used the two things I mentioned above.)
What I have done so far is
$ |\frac{n}{2}\int_{\mathbb{R}}e^{-n|t|} f(t)dt - 2 | = | \frac{n}{2}\int_{-\infty}^0 e^{nt} f(t)dt + \frac{n}{2}\int_{0}^{\infty} e^{-nt} f(t)dt - \frac{(3+1)}{2}| \\ \leq \frac{1}{2}|n\int_{-\infty}^0 e^{nt} f(t)dt - 3 | + \frac{1}{2}|n\int_{0}^{\infty} e^{-nt} f(t)dt-1|$
but here the issue is that I can't use $3 = 3\int_{-\infty}^0 e^{nt}dt$ and so on because, in order to use the positive summation kernel on a subinterval of $\mathbb{R}$, $t=0$ needs to be an interior point of that interval. Even if I could absorb it into a single integral, I'm not sure how I would use that limit. Those two limits have to do with letting $t$ approach 0 from each side, so you can see that each integral will be zero near the origin. But it has nothing to do with n, and it only holds for certain t:s anyway.
I'll show you how to do it for the expression $\left|n\int_{-\infty}^{0} e^{nt} f(t) \ dt - 3 \right|$ and you can do it for the other one. Let $\epsilon > 0$ be given. Since $\lim_{t \to 0^-} f(t) = 3$, it follows that: $$\exists \delta > 0: \forall t \in \mathbb{R}: 0 > t > -\delta \implies |f(t)-3| < \epsilon$$
Observe that: $$\int_{-\infty}^{0} ne^{nt} \ dt = 1$$ So, I can write $3 = 3\int_{-\infty}^{0} ne^{nt} \ dt = 1$. You can do this precisely because the integral above evaluates to $1$; it doesn't matter whether you call it a positive summation kernel or not. The sequence of functions $(K_n)$ is a positive summation kernel because it satisfies a list of properties that you're familiar with but, at the end of the day, they're just functions that you can manipulate like any other function. It follows that: $$\left|n \int_{-\infty}^{0} f(t) \ dt - 3 \right| \leq \int_{-\infty}^{0} ne^{nt} |f(t)-3| \ dt = \int_{-\delta}^{0} ne^{nt} |f(t)-3| \ dt + \int_{-\infty}^{-\delta} ne^{nt} |f(t)-3| \ dt$$
Let's deal with the second integral first. Since $f$ is bounded, I can always write: $$\int_{-\infty}^{-\delta} ne^{nt} |f(t)-3| \ dt \leq M \int_{-\infty}^{-\delta} ne^{nt} \ dt = M e^{-n\delta}$$ By picking a large enough $N \in \mathbb{N}$, you can show that when $n > N$, $Me^{-n\delta} < \epsilon$. So, the second integral above can be made really small by making $n$ very large. Next, let's deal with the first integral: $$\int_{-\delta}^{0} ne^{nt} |f(t)-3| \ dt \leq \int_{-\delta}^{0} \epsilon ne^{nt} \ dt = \epsilon (1-e^{-n\delta}) \leq \epsilon$$ which proves that for large enough $n$, $\lim_{n \to \infty} \left|n \int_{-\infty}^{0} e^{nt} f(t) \ dt - 3 \right| = 0$. For the other expression, the argument will come down to a similar bit of reasoning.