Almost everywhere equality and convolution.

Question

Firstly, let me say that my knowledge of measure theory is, at best, limited.

My question is, if you have four Lebesgue integrable functions that are pair-wise almost everywhere equal and you take the convolution of the "non-equal" functions, are resulting functions also almost everywhere equal?

In other words:

Let $f_1, f_2, g_1, g_2 : \mathbb{R} \to [0,1]$ be Lebesgue integrable functions such that $f_1 = g_1$ almost everywhere and $f_2 = g_2$ almost everywhere.

Does it hold that $(f_1 * f_2) = (g_1 * g_2)$ almost everywhere?

Any pointers on how to prove (or disprove) this will be helpfull and much appreciated.

Answer 1

The equality holds everywhere.

Observation: If $g_1=g_2\text{ a.e}$, then $f*g_1=f*g_2$ everywhere.

Proof: For all $x$, it holds $$\{y\in\Bbb R\,:\, f(x-y)g_1(y)\ne f(x-y)g_2(y)\}=\\=\{y\in\Bbb R\,:\, f(x-y)\ne 0\}\cap\{y\in\Bbb R\,:\, g_1(y)\ne g_2(y)\}\subseteq\\\subseteq \{y\in\Bbb R\,:\, g_1(y)\ne g_2(y)\}$$

So, for all $x$ the two functions $[y\mapsto f(x-y)g_1(y)]$ and $[y\mapsto f(x-y)g_2(y)]$ are equal almost-everywhere.

Therefore, they have the same integral over $\Bbb R$, i.e. $$(f*g_1)(x)=\int_{\Bbb R}f(x-y)g_1(y)\,dy=\int_{\Bbb R}f(x-y)g_2(y)\,dy=(f*g_2)(x)$$

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Using the observation above (and your notation, this time), you have these everywhere-equalities: $$g_1*g_2=f_1*g_2=f_1*f_2$$