= Solution
Fix $0<x<1$. On the open <interval> $(0,x)$, the <function> $y\mapsto(x-y)^{\beta-1}$ is finite, positive and continuous, and $f$ is nonnegative and <Lebesgue measurable>. Their product is therefore a nonnegative <measurable function>. Its <Lebesgue integral> is always defined as an element of $[0,\infty]$, with no possibility of cancellation of two infinities. Thus \b[$g(x)$ is well defined as a nonnegative extended integral].
The endpoint $y=x$ need not receive its formal power value: for $\beta<1$ that value would be singular. The integral is over $(0,x)$, and any chosen value at the endpoints has no effect because endpoints form a <null set>.
Finiteness is not implied by the present assumptions. For example, $f(y)=1/y$ is an admissible nonnegative <measurable function>. For any fixed $x$, the factor $(x-y)^{\beta-1}$ is bounded below by a positive constant on $0<y<x/2$, so $g(x)\geq C_x\int_0^{x/2}dy/y=\infty$. The later <L1 norm> bound will establish finiteness <almost everywhere> when $f$ is integrable. The construction is a <Volterra power-kernel integral>.
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