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.


1 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. $$(f*g_1)(x)=\int_{\Bbb R}f(x-y)g_1(y)\,dy=\int_{\Bbb R}f(x-y)g_2(y)\,dy=(f*g_2)(x)$$

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


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