Pointwise limit of convolution 
Suppose $\omega$ is the standard mollifier in $\mathbb R$. Then, let $\omega_{\epsilon} (x):= \frac{1}{\epsilon} \omega \left(\frac{x}{\epsilon}\right)$. For $0 < t_{1} < t_{2}$ the following result is to be proven:
  $$
\lim_{\epsilon \to 0} \int_{0}^{t} (\omega_{\epsilon}(s-t_1) - \omega_{\epsilon}(s-t_2))\,ds = \chi_{[t_1,t_2]}, 
$$
  where $\chi_{[t_1,t_2]}$ is the characteristic function on the interval $[t_1, t_2]$.

I couldn't prove the result. Also, I don't understand if the mode of convergence is pointwise or not.
What I have tried:
$$
\omega_{\epsilon}(s-t_1) - \omega_{\epsilon}(s-t_2) = \frac{1}{\epsilon} \left(\omega \left(\frac{s-t_1}{\epsilon}\right) - \omega \left(\frac{s-t_2}{\epsilon}\right)\right) = \frac{\epsilon}{t_2 - t_1} \omega' \left(\frac{s-t_1}{\epsilon} + \theta \frac{t_2 - t_1}{\epsilon}\right).
$$
So, 
$$
\int_{0}^{t} (\omega_{\epsilon}(s-t_1) - \omega_{\epsilon}(s-t_2))\,ds = \frac{1}{t_2 - t_1}\left[\omega\left(\frac{t-t_{1}}{\epsilon} + \theta \frac{t_2 - t_1}{\epsilon}\right) - \omega \left(\theta \frac{t_2}{\epsilon}+ \left(-\frac{1+\theta}{\epsilon}\right)t_1\right)\right].
$$
But then, I have lost my way regarding "letting $\epsilon \to 0$" . Please help!
 A: You are thinking too complicated. What we need is that the integral of the mollifier is $1$, and that the support of $\omega_\epsilon$ shrinks to the origin as $\varepsilon \to 0$ (if we took more general mollifiers, where the support need not be compact, we would still have $\lim\limits_{\epsilon \to 0} \int_{\lvert x\rvert > \delta} \lvert\omega_\epsilon(x)\rvert\,dx = 0$ for every $\delta > 0$, and that would suffice), and for a certain point, we need the symmetry $\omega(-x) = \omega(x)$ (allowing non-symmetric mollifiers wouldn't change anything essential, however).
Fix an arbitrary $t\in \mathbb{R}$, and consider
$$\int_0^t \omega_\epsilon(s - t_1)\,ds.$$
It is easier to see what happens if we have $\omega_\epsilon(u)$ as the integrand, so let's make the substitution $u = s - t_1$. We obtain
$$\int_{-t_1}^{t-t_1} \omega_\epsilon(u)\,du.\tag{1}$$
Now look at
$$\lim_{\epsilon \to 0} \int_a^b \omega_\epsilon(u)\,du$$
for arbitrary $a,b\in \mathbb{R}$. If $a = b$, the integral is clearly $0$ for all $\epsilon$, and if $b < a$, flipping the integral limits gives a sign change, so we may assume $a < b$.
If $b < 0$, or $0 < a$, then for all small enough $\epsilon$ the support of $\omega_\epsilon$ - which is $[-\epsilon, +\epsilon]$ for the mollifier used here - does not intersect the interval of integration. Then the integrand vanishes identically on $[a,b]$, and we have $\int_a^b \omega_\epsilon(u)\,du = 0$ for all small enough $\epsilon$.
If $a < 0 < b$, then for all small enough $\epsilon$ the support of $\omega_\epsilon$ is contained in the interval $[a,b]$, and then
$$\int_a^b \omega_\epsilon(u)\,du = \int_{-\infty}^{+\infty} \omega_\epsilon(u)\,du = 1.$$
If $a = 0$ or $b = 0$, then for all small enough $\epsilon$ the support of $\omega_\epsilon$ is contained in $(-\infty, b]$ resp. $[a,+\infty)$, and we have
$$\int_a^b \omega_\epsilon(u)\,du = \int_0^{+\infty} \omega_\epsilon(u)\,du = \frac{1}{2}$$
or
$$\int_a^b \omega_\epsilon(u)\,du = \int_{-\infty}^0 \omega_\epsilon(u)\,du = \frac{1}{2},$$
where the value $\frac{1}{2}$ comes from the symmetry $\omega(-x) = \omega(x)$. So we have
$$\lim_{\epsilon \to 0} \int_a^b \omega_\epsilon(u)\,du = \begin{cases} 0 &, 0 \notin [a,b] \\ 1 &, 0 \in (a,b) \\ \frac{1}{2} &, a = 0 \text{ or } b = 0.\end{cases}\tag{2}$$
Since by assumption $t_1 > 0$, we obtain
$$\lim_{\epsilon \to 0} \int_{-t_1}^{t-t_1} \omega_\epsilon(u)\,du = \begin{cases} 0 &, t < t_1 \\ 1 &, t > t_1 \\ \frac{1}{2} &, t = t_1 \end{cases}\tag{3}$$
when using the integral limits of $(1)$ in $(2)$.
The case for $\int_0^t \omega_\epsilon(s - t_2)\,ds$ is completely analogous, and so overall we have
$$\lim_{\epsilon \to 0} \int_0^t \omega_\epsilon(s-t_1) - \omega_\epsilon(s-t_2)\,ds = \begin{cases} 0 - 0 &, t < t_1 \\ \frac{1}{2} - 0 &, t = t_1 \\ 1 - 0 &, t_1 < t < t_2 \\ 1 - \frac{1}{2} &, t = t_2 \\ 1 - 1 &, t > t_2.\end{cases}$$
Thus the pointwise limit is not quite $\chi_{[t_1,t_2]}$, at the end points of the interval the limit value is $\frac{1}{2}$ rather than $1$. Pointwise, the limit function is $\chi_{[t_1,t_2]} - \frac{1}{2} \chi_{\{t_1,t_2\}} = \chi_{(t_1,t_2)} + \frac{1}{2} \chi_{\{t_1,t_2\}}$. But as far as integration with respect to the Lebesgue measure is concerned, all of these are indistinguishable, and we have
$$\lim_{\epsilon \to 0} \int_0^t \omega_\epsilon(s-t_1) - \omega_\epsilon(s-t_2)\,ds = \chi_{[t_1,t_2]}(t)$$


*

*in measure,

*pointwise almost everywhere,

*in $L^p(\mathbb{R})$ for $1 \leqslant p < \infty$.


We also have locally uniform convergence on $\mathbb{R} \setminus \{t_1,t_2\}$. But we don't have convergence in $L^\infty(\mathbb{R})$, since for every $\epsilon > 0$ the set
$$\biggl\{ t \in \mathbb{R} : \biggl\lvert \int_0^t \omega_\epsilon(s-t_1) - \omega_\epsilon(s-t_2)\,ds - \chi_{[t_1,t_2]}(t)\biggr\rvert > \frac{1}{4}\biggr\}$$
is a nonempty open set, and therefore has positive measure.
