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if $\int g\phi =\int f\phi$ for all $\phi \in C^{\infty}_{c}(U)$, then $f=g \quad a.e.$

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In Folland's real analysis textbook, on page 283, he mentioned that for an open set $U \subset \mathbb{R}^{n}$, and $f,g \in L_{loc}^{1}(U)$ .$f$ and $g$ define the same distribution precisely when they are equal a.e.

I wonder how do we show this formally, that is if $\int g\phi =\int f\phi$ for all $\phi \in C^{\infty}_{c}(U)$, then $f=g \quad a.e.$

Intuitively, my rough proof idea is to use contradiction and partition of unity, where in order to use a partition of unity, we need some open set containing a compact set structure to build a bump function. But I'm having trouble pinning down the exact detail, maybe something like $f_{n} \to f$ uniformly on some compact set and $g_{n} \to g$ uniformaly on some compact set, and $|f_{n}-g_{n}|>\epsilon$ on some rectangle and then build the bump function of rectangle. However, I'm not certain how to exactly do so.


Here seems to the answer, I voted to close this, but if there are other ways to attack this problem, might be interesting to see as well.


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