If a function is equal to an integrable function almost everywhere, is it integrable?

Question

Let $f:[a,b]\to\mathbb{R}$ be a Lebesuge-integrable function, and $g:[a,b]\to\mathbb{R}$ a function such that $f=g$ almost everywhere. I can prove that if $g$ is Lebesgue-measurable, then $g$ is Lebesgue-integrable. But is the condition that $g$ is Lebesgue-measurable necessary?

Answer 1

Thanks to @Kavi Rama Murthy and @José Carlos Santos's hints, I looked up the book I was reading (Berberian's Fundamentals of Real Analysis) and found a proof (really an exercise problem with a hint) that in a complete measure space $X$, if $f=g$ a.e. and $f$ is measurable, then $g$ is measurable:

Let $N\subset X$ be a null set outside of which $f=g$. Let $c\in\mathbb{R}$, and $$A=\{x\in X\mid f(x)>c\}\qquad\mbox{and}\qquad B=\{x\in X\mid g(x)>c\}.$$ Since $f$ is measurable, $A$ is measurable. We will show that $B$ is measurable. Clearly $A-B\subset N$ and $B-A\subset N$. Since $X$ is complete, $A-B$ and $B-A$ are measurable. Then $$B=(A-(A-B))\cup(B-A)$$ is measurable.

Thanks guys! (I have to say that the proof is not at all trivial.)