For AES CBC, can I encrypt the IV with AES ECB and the same key and include it with the message? Is it safe to use this process:


Generate Key, IV
Encrypt Message with Key, IV and AES CBC into EMessage 
Encrypt IV with ECB into EIV
Send over EIV + EMessage


Then the recipient


Decrypt EIV with Key into IV
Decrypt EMessage with IV and Key into Message


The thing that I'm worried about is that encrypting the IV with the same key gives additional information about the first block (and therefore subsequent blocks).

Edit: There was a great response with a direct proof in the case that the first block is all zeroes. What about this revised process:


Generate Key, IV1, IV2
Encrypt IV2 + Message with Key, IV and AES CBC into EMessage
Encrypt IV with ECB into EIV
Send over EIV + EMessage


Decrypt: 


Decrypt EIV with Key into IV
Decrypt EMessage with Key and IV into IV2 + Message
Throw away IV2


This seems to counter the information lost at the cost of one additional block of data in the message.
It actually leaks information. You are sending:


Encrypted IV: $AES(k,IV)$
First ciphertext block of CBC: $AES(k, M_1 \oplus IV)$


Eavesdropper can observe whether the two blocks are equal, which happens iff $M_1$ is all zeroes.