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No: the unbounded length of the domain matters. Let $\varphi (t)=\max \{1-|t|,0\}$ and define $f_n(x)=n^{-1}\varphi (x/n)$ on $\mathbb{R}$. These are continuous functions with compact support, and ${\sup }_{\mathbb{R}}|f_n|=1/n\to 0$, so the sequence has uniform convergence to zero. However, the change of variables $t=x/n$ givesThe integral of the limit is zero. The supports widen as the heights decrease; a uniform bound on the heights gives no bound on the total area over an infinite interval.

The functions $g_n$ are nonnegative. The change of variables $u=nx$ and the given Gaussian integral implyFor every fixed $\delta >0$,Thus the mass concentrates at the endpoint zero. This is a one-sided approximate identity.

Because $f$ is continuous on the compact interval $[0,1]$, it is a bounded function; put $B={\sup }_{[0,1]}|f|$. Given $\epsilon >0$, choose $\delta \in (0,1)$ with $|f(x)-f(0)|<\epsilon$ for $0\le x<\delta$. Splitting the integral near and away from zero yieldsTaking the limit superior gives a bound of $\epsilon$, and $\epsilon$ was arbitrary. ConsequentlyOnly continuity at zero and boundedness on $[0,1]$ were needed for this final estimate; the assumed continuity everywhere also guarantees the ordinary integrals exist.