Let $f$ and $g$ be defined on $R$ by
$f(x)$ is Dirichilet's function another one is $g(x) = 1$ if $x\geq 0$ and $0$ elsewhere. Which of the following statements are true?
a. The function $f$ is continuous almost everywhere.
b. The function $f$ is equal to a continuous function almost everywhere.
c.The function $g$ is equal to a continuous function almost everywhere.
Can anyone help me how to proceed?
On
The other answer should tackle the other potentially tricky one.
For $c$, maybe try drawing it. Notice that any continuous function equal to $g$ almost everywhere would have to climb infinitely fast, which is not possible, or climb from $0$ to $1$ on a set of nonzero measure.
If you have seen sequences of functions, it might also be helpful to note that $g$ is the pointwise limit of the sequence $$ f_n(x)=\begin{cases}0&x\leq-\frac1n\\ nx+1& -\frac{1}{n}\leq x\leq 0 \\ 1&x\geq0\end{cases} $$ Each function is continuous, can get as close as you want (indeed the set that they aren't identical is shrinking to $0$), but you need to take a limit to make sure that set is actually negligible. So we can't approximate this in maybe the most "natural" way.
By way of proof, suppose we could with some function $f$. Then, for any interval we should have $$ \int_a^b|f(x)-g(x)|\mathrm dx=0 $$ Then, we have some $\delta>0$ with $$ |x|<\delta\implies \frac{1}{2}<f(x) $$ and $$ \int_{-\delta}^0|f(x)-g(x)|\mathrm dx\\ =\int_{-\delta}^0|f(x)|\mathrm dx>\delta/2>0 $$
The choice $b$ is true since the Dirichlet function is equal to $0$ off $\mathbb{Q}$ and $\mathbb{Q}$ has measure zero. The rest are false. I'll leave those to you.