Convolution and integration Let $f\in L_\text{loc}^1(\Omega)$ be such that $f \ge0$, and $\int_\Omega f\varphi ~dx=0$ for all $\varphi\in \mathcal{C}^\infty_C(\Omega)$. Show that $f=0$ a.e. in $\Omega$.
i am not able to get how to approach for it. Any type of help will be appreciated. Thanks in advance.
 A: There is a bare hands approach to this which is similar to showing that $C^{\infty}_c(\Omega)$ is dense in $L^1(\Omega)$:
Roughly speaking: Suppose that $f > 0$ on a set $E$ of positive measure. Now take $\phi$ to be a very good approximation of $\mathbf{1}_E$.
A: To reply to your comment...
The idea is to show that $$\int f 1_{B(x, r)} = 0$$ for all open balls $B(x,r) \subset  \Omega$. Once we have done this, the Lebesgue differentiation theorem tells us that
$$ f(x) = \lim_{r\to 0} \tfrac{1}{{\rm Vol}(B(x,r))} \int_{B(x,r)} f  = 0$$
for almost every $x$, which is what we want to prove.
To show that $\int f 1_{B(x, r)} = 0$ for all open balls, we want to approximate each $1_{B(x,r)}$ using a sequence $\varphi_n$ of smooth functions. We need that ensure that


*

*Each $\varphi_n$ is smooth and compactly supported (so that $\int f \varphi_n = 0$ for each $n$, by assumption);

*$\varphi_n $ converges to $ 1_{B(x,r)}$ pointwise as $n \to \infty$;

*$\varphi_n \leq 1_{B(x,r)}$ for all $n \in \mathbb N$ (so that we can apply Dominated Convergence).
Once we have found such a sequence $\varphi_n$, we can deduce that
$$ \int f 1_{B(x,r)} = \int f \lim_{n \to \infty} \varphi_n = \lim_{n\to \infty} \int f \varphi_n = 0,$$
which is what we want.

To keep notation simple, I'll deal with the unit open ball $B(0,1)$ centred at the origin, and I'll work in one dimension. Let us begin by considering this function:
$$ \eta_n(t) = \begin{cases} \exp\left[ - \frac 1 {\left(t-(1-\tfrac 1 n)^2\right)(1-t)}\right] & {\rm if \ \ } t \in ((1- \tfrac 1 n)^2, 1) \\ 0  & {\rm otherwise} \end{cases}.$$
This function looks like a "bump": it is smooth, and is supported on the interval $[(1-\tfrac 1 n)^2, 1]$.
Now, consider the function
$$ \varphi_n (x) = \frac{\int_{x^2}^\infty \eta_n (t) dt}{\int_{-\infty}^\infty \eta_n(t) dt}.$$
This function is also smooth (by the fundamental theorem of calculus). Furthermore, you can verify that
$$ |x| \leq 1- \tfrac 1 n\implies \varphi_n(x) = 1,$$
and
$$ |x| \geq 1 \implies \varphi_n(x) = 0.$$
So $\varphi_n$ is another "bump function", this time, supported on $[-1,1]$.
Furthermore, $\varphi_n$ satisfies an additional nice property which is that it is constant and identically equal to one on the entire interval $[-1+\tfrac 1 n, 1-\tfrac 1 n]$.
Finally, on the intervals $[-1,-1 + \tfrac 1 n]$ and $[1 - \tfrac 1 n , 1]$, $\varphi_n$ smoothly interpolates between the values of zero and one.
By construction, each $\varphi_n$ is smooth and compactly supported.  $\varphi_n$ converges pointwise to $1_{B(0,1)}$ as $n \to \infty$. And $\varphi_n \leq 1_{B(0,1)}$ for all $n$. So the sequence $\varphi_n$ satisfies all the desired properties.
