Don't leave it unproved .

Total number of 5 digits are 90000.So probability suggests that the chance for the sum of the digits in a number are equal for both odd and even, which is 1/2.So 90000/2 are the 5 digit numbers whose digits' sum is even. THis is the easiest logical method.

$9*({10}^{4}) - ({5}^{5}*\left( \begin{matrix} 4 \\ 0 \end{matrix} \right) + \left( \begin{matrix} 4 \\ 1 \end{matrix} \right)*({5}^{3}*4*5) + {5}^{5}*\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) + \left( \begin{matrix} 4 \\ 3 \end{matrix} \right)*({5}^{3}*4*5) + {5}^{5}\left( \begin{matrix} 4 \\ 4 \end{matrix} \right))$

How I reached to it is quite understandable, I guess.

That is what is not actually a nice solution. U are just taking half as a nice choice. Try to use combinatorics and finally take out 45000. Prove half will be solutionb

I have first finded out that total 5 digit numbers is 90000 and out of which half are even and other half are no other possibilities. Thus total even 5 digit numbers are 90000/2 .
Thanks @Utsav Singhal, as I havn tried this for first time. Yeah :).