A smooth, compactly supported, normalized, positive, etc. function is called mollifier if: $$\varphi_\varepsilon(x):=\frac{1}{\varepsilon}\varphi\left(\frac{x}{\varepsilon}\right):\quad\varphi_\varepsilon(x)\to\delta(x)$$ (See wikipedia: Mollifiers: Definition )
Now I'm just wondering wether this requirement can fail in general: $$\varphi\in\mathcal{C}_c^\infty(\mathbb{R}):\quad\varphi_\varepsilon(x)\nrightarrow\delta(x)\quad\left(\int\varphi(x)dx=1\right)$$ (I guess so but what would be an appropriate example?)
The condition is redundant. In fact, we have the stronger
Proposition: Let $\varphi \in L^1(\mathbb{R}^n)$ with $\int_{\mathbb{R}^n} \varphi(x)\,dx = 1$. For $\varepsilon > 0$ define $\varphi_\varepsilon \colon x \mapsto \varepsilon^{-n}\cdot \varphi(x/\varepsilon)$. Then
$$\lim_{\varepsilon \searrow 0} \varphi_\varepsilon = \delta$$
in the sense of distributions.
Proof: Let $\psi \in \mathscr{D}(\mathbb{R}^n)$. Then
\begin{align} \left\lvert \int_{\mathbb{R}^n}\varphi_{\varepsilon}(x)\psi(x)\,dx - \psi(0)\right\rvert &= \left\lvert \int_{\mathbb{R}^n} \varphi(y)\psi(\varepsilon y)\,dy - \psi(0)\right\rvert\\ &= \left\lvert \int_{\mathbb{R}^n}\varphi(y)\bigl(\psi(\varepsilon y) - \psi(0)\bigr)\,dy\right\rvert\\ &\leqslant \int_{\mathbb{R}^n} \lvert \varphi(y)\rvert\cdot \lvert \psi(\varepsilon y) - \psi(0)\rvert\,dy\\ &= \int_{\lvert y\rvert \leqslant R} \lvert \varphi(y)\rvert\cdot \lvert \psi(\varepsilon y) - \psi(0)\rvert\,dy + \int_{\lvert y\rvert > R} \lvert \varphi(y)\rvert\cdot \lvert \psi(\varepsilon y) - \psi(0)\rvert\,dy\\ &\leqslant \sup_{\lvert z\rvert \leqslant \varepsilon R} \lvert \psi(z) - \psi(0)\rvert \int_{\lvert y\rvert \leqslant R} \lvert\varphi(y)\rvert\,dy + 2\max_{z\in \mathbb{R}^n} \lvert\psi(z)\rvert \int_{\lvert y\rvert > R} \lvert\varphi(y)\rvert\,dy\\ &\leqslant \lVert\varphi\rVert_{L^1} \cdot \lVert\psi\rVert_{H^{1,\infty}}\cdot R\cdot \varepsilon + 2\lVert \psi\rVert_{L^\infty} \underbrace{\int_{\lvert y\rvert > R} \lvert\varphi(y)\rvert\,dy}_{A(R)}. \end{align}
Since $\varphi \in L^1(\mathbb{R}^n)$, we have $A(R) \to 0$ as $R\to \infty$, and thus, for any given $\eta > 0$, we can choose $R\in (0,\infty)$ such that
$$2\lVert \psi\rVert_{L^\infty} \cdot A(R) < \frac{\eta}{2}.$$
Then we can set
$$\varepsilon_0 := \frac{\eta}{1 + 2\lVert\varphi\rVert_{L^1}\lVert\psi\rVert_{H^{1,\infty}}R}$$
and have
$$\left\lvert \int_{\mathbb{R}^n} \varphi_\varepsilon(x)\psi(x)\,dx - \psi(0) \right\rvert < \eta$$
for all $\varepsilon \leqslant \varepsilon_0$. The choices of $R$ and $\varepsilon_0$ can be made uniformly on bounded subsets of $\mathscr{D}(\mathbb{R}^n)$ (just replace the $\lVert\psi\rVert_{X}$ expressions by positive constants), hence we have
$$\varphi_\varepsilon \to \delta$$
in the sense of distributions.